K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

25 tháng 7 2023

Giúp mình vớiiii

 

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

b: \(=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}\)

16 tháng 5 2021

`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

16 tháng 5 2021

`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

15 tháng 7 2021

`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`

`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`

`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`

`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`

`=1/(sqrtx-2)`

15 tháng 7 2021

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

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

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

a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

\(=a+\sqrt{a}-2\sqrt{a}+1+1\)

\(=a-\sqrt{a}+2\)

28 tháng 8 2021

a,ĐKXĐ: tự tìm :v

 \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)

\(A=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

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

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

12 tháng 7 2021

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(A< \dfrac{3}{5}\Rightarrow\dfrac{3}{5}-A>0\Rightarrow\dfrac{3}{5}-\dfrac{\sqrt{x}-3}{\sqrt{x}-1}>0\)

\(\Rightarrow\dfrac{3\left(\sqrt{x}-1\right)-5\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}>0\Rightarrow\dfrac{12-2\sqrt{x}}{5\left(\sqrt{x}-1\right)}>0\)

\(\Rightarrow\dfrac{2}{5}.\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\Rightarrow\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}6-\sqrt{x}< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1< x< 36\\\left\{{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\left(l\right)\end{matrix}\right.\) 

\(\Rightarrow1< x< 36\)

 

12 tháng 7 2021

\(=>A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

để \(A< \dfrac{3}{5}< =>\dfrac{\sqrt{x}-3}{\sqrt{x}-1}< \dfrac{3}{5}\)

\(< =>\dfrac{5\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{5\left(\sqrt{x}-1\right)}< 0\)

\(< =>\dfrac{2\sqrt{x}-12}{5\left(\sqrt{x}-1\right)}< 0\)

\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}2\sqrt{x}-12>0\\5\left(\sqrt{x}-1\right)< 0\end{matrix}\right.\\\left[{}\begin{matrix}2\sqrt{x}-12< 0\\5\left(\sqrt{x}-1\right)>0\end{matrix}\right.\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\\\left[{}\begin{matrix}x< 36\\x>1\end{matrix}\right.\end{matrix}\right.=>1< x< 36\left(tm\right)\)

AH
Akai Haruma
Giáo viên
1 tháng 4 2021

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1; x\neq 25$

a) 

\(A=\frac{4\sqrt{x}}{\sqrt{x}-5}:\left[\frac{(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2}+\frac{5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\right]\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}}{\sqrt{x}-5}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{4(\sqrt{x}+2)}{\sqrt{x}-5}\)

b) Tại $x=81$ thì $\sqrt{x}=9$.

Khi đó: $A=\frac{4(9+2)}{9-5}=11$

c) $A< 4\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}-5}< 1$

$\Leftrightarrow \frac{7}{\sqrt{x}-5}< 0\Leftrightarrow \sqrt{x}-5< 0$

$\Leftrightarrow 0\leq x< 25$. Kết hợp với ĐKXĐ suy ra: $0\leq x< 25; x\neq 1$

1 tháng 4 2021

Hỗ trợ em nhé cô

Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x-\sqrt{x}}\right)\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)

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

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

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)