A= $\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a} 3}\right)$ ĐK:(a≥0, a≠9) B= \(\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a} ...

3

NT

Nguyễn Tấn An

15 tháng 8 2018

$A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}$$B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}$

NT

Nguyễn Tấn An

16 tháng 8 2018

$C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a$$D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}$

A= $\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)$ ĐK: (a≥0, a≠1) B= $\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}$ ĐK: (a>0, a≠0, a≠2) C= $\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)$ ĐK: (a>0, a≠1) D= $\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}$ ĐK: (a>0) E=...

QC

Qúy Công Tử

25 tháng 7 2018

1

AT

Aki Tsuki

12 tháng 8 2018

A = $\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-2\right)\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right)=\left(\sqrt{a}+1-2\right)\left(\sqrt{a}+1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)=a-1$

$B=\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}=\left(\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}=\left(\dfrac{a}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{\left(\sqrt{a}-1\right)\left(a-2\right)}{\sqrt{a}\left(a+2\right)}$

$C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\dfrac{a}{a-1}\right):\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)-\sqrt{a}}{\sqrt{a}+1}\right)=\dfrac{\sqrt{a}}{a-1}:\dfrac{a}{\sqrt{a}+1}=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{a}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}$

$D=\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}=\sqrt{a}+1+\dfrac{a+4}{\sqrt{a}+2}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{a+2\sqrt{a}+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{2a+3\sqrt{a}+6}{\sqrt{a}+2}$

$E=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a-1+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\cdot\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}\cdot\sqrt{a}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}$

MN

minh ngọc

23 tháng 8 2023

$\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)$
$\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}$
RÚT GONJ

2

H

HaNa

23 tháng 8 2023

2)

ĐK: $x\ge0;x\ne4$

Biểu thức trở thành:

$\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+2\sqrt{a}+3\sqrt{a}+6}{a-4}-\dfrac{a-2\sqrt{a}-\sqrt{a}+2}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\\ =\dfrac{4\sqrt{a}+8}{a-4}\\ =\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\\ =\dfrac{4}{\sqrt{a}-2}$

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NL

Nguyễn Lê Phước Thịnh

24 tháng 8 2023

1:

$\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)$

$=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}$

$=\dfrac{x+2\sqrt{x}-8-x-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}$

$=\dfrac{-2\sqrt{x}-11}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(2\sqrt{x}+11\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-3\right)}$

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N

nchdtt

7 tháng 7 2021

$A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}$

$B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}$ (ĐK x>0; x$\ne9$)

a)Rút gọn A và B

b) Tìm các giá trị của x để giá trị biểu thức A lớn hơn giá trị biểu thức B

1

AT

An Thy

7 tháng 7 2021

a) $A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}$

$=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|$

$=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}$

$B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}$

$=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}$

$=\dfrac{8}{\sqrt{x}-3}$

b) $A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0$

$\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0$

Ta có: $\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0$

$\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9$

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TM

Trịnh Minh Tuấn

19 tháng 9 2021

$A=\left(\dfrac{2a+1}{\sqrt{a^3}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)$(đk a lớn hơn bằng 0,a khác 1)

a, rút gọn a

b,tìm a để A=6

Rút gọn: A = $\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}$ B = $\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18$ C = $\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}$với n thuộc N* D =...

1

LL

Lấp La Lấp Lánh

19 tháng 9 2021

a) $A=\left(\dfrac{2a+1}{\sqrt{a^3}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\left(đk:a\ge0,a\ne1\right)$

$=\dfrac{2a+1-\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}.\left[\dfrac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right]$

$=\dfrac{2a+1-a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}.\left(a-\sqrt{a}+1-\sqrt{a}\right)$

$=\dfrac{a+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}.\left(\sqrt{a}-1\right)^2$

$=\sqrt{a}-1$

b) $A=\sqrt{a}-1=6$

$\Leftrightarrow\sqrt{a}=7\Leftrightarrow a=49$

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PT

Phương Thảo

7 tháng 7 2017

Bài 75 (trang 40 SGK Toán 9 Tập 1) Chứng minh các đẳng thức sau: a) $\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=-1,5$; b) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=-2$; c) $\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b$ với $a, b$ dương và $a \neq b$; d)...

0

KK

Kim Khánh Linh

25 tháng 4 2021

111

NA

Nguyễn Anh Tuấn

17 tháng 5 2021

) V T = ( 2 √ 3 − √ 6 √ 8 − 2 − √ 216 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 2 ⋅ √ 3 − √ 6 √ 2 2 ⋅ 2 − 2 − √ 6 2 .6 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 6 − √ 6 2 √ 2 − 2 − 6 . √ 6 3 ) ⋅ 1 √ 6 = [ √ 6 ( √ 2 − 1 ) 2 ( √ 2 − 1 ) − 6 √ 6 3 ] ⋅ 1 √ 6 = ( √ 6 2 − 2 √ 6 ) ⋅ 1 √ 6 = ( √ 6 2 − 4 √ 6 2 ) ⋅ 1 √ 6 = ( − 3 2 √ 6 ) ⋅ 1 √ 6 = − 3 2 = − 1 , 5 = V P . b) V T = ( √ 14 − √ 7 1 − √ 2 + √ 15 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = ( √ 7 ⋅ √ 2 − √ 7 1 − √ 2 + √ 5 ⋅ √ 3 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = [ √ 7 ( √ 2 − 1 ) 1 − √ 2 + √ 5 ( √ 3 − 1 ) 1 − √ 3 ] : 1 √ 7 − √ 5 = ( − √ 7 − √ 5 ) ( √ 7 − √ 5 ) = − ( √ 7 + √ 5 ) ( √ 7 − √ 5 ) = − ( 7 − 5 ) = − 2 = V P . c) V T = a √ b + b √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a ⋅ √ b + √ b ⋅ √ b ⋅ √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a b + √ b ⋅ √ a b √ a b : 1 √ a − √ b = √ a b ( √ a + √ b ) √ a b ⋅ ( √ a − √ b ) = ( √ a + √ b ) ⋅ ( √ a − √ b ) = a − b = V P . d) V T = ( 1 + a + √ a √ a + 1 ) ( 1 − a − √ a √ a − 1 ) = ( 1 + √ a ⋅ √ a + √ a √ a + 1 ) ( 1 − √ a ⋅ √ a − √ a √ a − 1 ) = [ 1 + √ a ( √ a + 1 ) √ a + 1 ] [ 1 − √ a ( √ a − 1 ) √ a − 1 ] = ( 1 + √ a ) ( 1 − √ a ) = 1 − ( √ a ) 2 = 1 − a = V P

NT

Nguyễn Trung Kiên

19 tháng 5 2021

a) $VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4 \sqrt{6}}{2}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{-3}{2} \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=-\dfrac{3}{2}=-1,5=V P$ .
b) $VT=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$=\left(\dfrac{\sqrt{7} \cdot \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5} \cdot \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$=\left[\dfrac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]: \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$sqrt{7}-\sqrt{5})$

$= - (7 - 5) = - 2 = V P$ .

c) $T=\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b}+\sqrt{b} \cdot \sqrt{b} \cdot \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a} \cdot \sqrt{a b}+\sqrt{b} \cdot \sqrt{a b}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})$

$=(\sqrt{a}+\sqrt{b}) \cdot(\sqrt{a}-\sqrt{b})$

$= a - b = V P$ .

d) $VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)$

$left(1-\dfrac{\sqrt{a} \cdot \sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\right)$

$left[1-\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]$

$1-\sqrt{a})$

$=1-(\sqrt{a})^{2}=1-a=V P$

P

pansak9

12 tháng 9 2023

Cho biểu thức: $A$ = $\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}-\dfrac{4x}{x-9}\right)$ : $\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{3\sqrt{x}-x}\right)$ . Tìm đk của x để |A| > - A

1

NL

Nguyễn Lê Phước Thịnh

12 tháng 9 2023

ĐKXĐ: x>0; x<>9

$A=\left(\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{4x}{x-9}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)$

$=\dfrac{-x-6\sqrt{x}-9+x-6\sqrt{x}+9-4x}{x-9}:\dfrac{-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-3\right)}$

$=\dfrac{-4x-12\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-2\right)}$

$=\dfrac{4x\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}-2\right)}=\dfrac{4x}{\sqrt{x}-2}$

|A|>-A

=>A>=0

=>4x>0

=>x>0 và x<>9

LN

Lương Ngọc Anh

29 tháng 6 2023

28. A=$\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)$
a. rút gọn A
b. tính A với x = $7-4\sqrt{3}$
c. tìm x khi A=3

2

NL

Nguyễn Lê Phước Thịnh

29 tháng 6 2023

a:

$A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}$

$=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}$

b: Khi x=7-4căn 3 thì

$A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}$

c: A=3

=>căn x-2=1

=>x=9(loại)

Y

YangSu

29 tháng 6 2023

$a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)$

$=\dfrac{x-3\sqrt{x}-x+9}{x-9}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}$

$=\dfrac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{9-x+x-9-x+4\sqrt{x}-4}$

$=\dfrac{-3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-2}{4\sqrt{x}-4-x}$

$=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(x-4\sqrt{x}+4\right)}$

$=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}$

$=\dfrac{3}{\sqrt{x}-2}$

$b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}$

$c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)$

Vậy không có giá trị x thỏa mãn đề bài.

Đúng(1)

T

títtt

11 tháng 1

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