Rút gọn biểu thứca)$\sqrt{\left(x 1\right)^2}$$\left(x\ge-1\right)$b)$\sqrt{\left(2x-1\right)^4}$c)$\sqrt{\left(x-1\right)^2}$$\left(x>1\right)$d) $\sqrt{x^2-4x 4}$$\left(x\ge2\right)$e)...

0

MT

Mai Thị Thúy

22 tháng 7 2021

giải pt :

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

b, $\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21$

c, $\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9$

d, $\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}$

#Toán lớp 11

0

HK

Hoàng Kiều Quỳnh Anh

22 tháng 3 2022

Rút gọn biểu thức

a) A=$2\sqrt{\left(2-\sqrt{5}\right)^2}-\dfrac{8}{3-\sqrt{5}}$

b) B= $\left(\dfrac{2\sqrt{x}}{x-4}-\dfrac{1}{\sqrt{x}+2}\right):\left(1+\dfrac{2}{\sqrt{x}-2}\right)$ Với x>0, x khác 4

1

NV

Nguyễn Việt Lâm

Giáo viên

22 tháng 3 2022

$A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}$

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

$=2\sqrt{5}-4-6-2\sqrt{5}=-10$

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

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

$=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}$

6M

6.Phạm Minh Châu

9 tháng 10 2021

Câu 1: Rút gọn biểu thức sau:

a.$\sqrt{36\left(x-5\right)^2}$

b. $\sqrt{\dfrac{1}{4}\left(1-x\right)^2}$

c.$\sqrt{x^2\left(2x-4\right)^2}$

1

KV

Kiều Vũ Linh

9 tháng 10 2021

a) $\sqrt{36\left(x-5\right)^2}=6\left|x-5\right|$

$=6\left(x-5\right)$ (khi $x\ge5$)

hoặc $=6\left(5-x\right)$ (khi $x< 5$)

b) $\sqrt{\dfrac{1}{4}\left(1-x\right)^2}=\dfrac{1}{2}\left|1-x\right|$

$=\dfrac{1}{2}\left(1-x\right)$ (khi $x\le1$)

hoặc $=\dfrac{1}{2}\left(x-1\right)$ (khi $x>1$)

c) $\sqrt{x^2\left(2x-4\right)^2}=\left|x\right|\left|2x-4\right|$

$=x\left(2x-4\right)$ (khi $x\ge2$)

hoặc $=x\left(4-2x\right)$ (khi $0\le x< 2$)

hoặc $=-x\left(4-2x\right)$ (khi $x< 0$)

Đúng(1)

AD

An Đinh Khánh

16 tháng 8 2023

Rút gọn biểu thức sau
A=$\dfrac{1}{x-1}\sqrt{75\left(x-1\right)^3}\left(x>1\right) $
B=$5\sqrt{4x}-3\sqrt{\dfrac{100x}{9}}-\dfrac{4}{x}\sqrt{\dfrac{x^3}{4}}\left(x>0\right) $
C=$x-4+\sqrt{16-8x+x^2}\left(x>4\right)$
Help me

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 8 2023

a: $A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}$

$=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}$

b: $B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}$

$=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}$

c: $C=x-4+\left|x-4\right|$

=x-4+x-4

=2x-8

giải pt :a,$\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)$ b, $\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21$ c, $\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9$ ...

MT

Mai Thị Thúy

22 tháng 7 2021

#Toán lớp 11

1

MT

Mai Thị Thúy

22 tháng 7 2021

mong mọi người giải giúp em vs

NN

Nguyễn Nhã Thanh

11 tháng 7 2017

Rút gọn các biểu thức sau:
a)$x+3+\sqrt{x^2-6x+9}\left(x\le3\right)$
b) $\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)$
c) $\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)$
d) $\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\left(x< 2\right)$

1

LD

l҉o҉n҉g҉ d҉z҉

26 tháng 8 2020

a) $x+3+\sqrt{x^2-6x+9}\left(x\le3\right)$

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

$=x+3+\left|x-3\right|$

$=x+3-\left(x-3\right)$

$=x+3-x+3$

$=6$

b) $\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)$

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

$=\left|x+2\right|-\left|x\right|$

$=x+2-\left(-x\right)$

$=x+2+x$

$=2x+2=2\left(x+1\right)$

c) $\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)$

$=\frac{\sqrt{\left(x-1\right)^2}}{x-1}$

$=\frac{\left|x-1\right|}{x-1}$

$=\frac{x-1}{x-1}=1$

d) $\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}$

$=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}$

$=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}$

$=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}$

$=\left|x-2\right|-1$

$=-\left(x-2\right)-1$

$=-x+2-1$

$=-x+1=-\left(x-1\right)$

DT

Duong Tue Tam

16 tháng 6 2023

Rút gọn các biểu thức...

1

Y

YangSu

16 tháng 6 2023

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

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

$=3x+6\sqrt{x}-x+1$

$=2x+6\sqrt{x}+1$

$B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2$

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

$=x+4\sqrt{x}+3-2x+4\sqrt{x}-2$

$=-x+8\sqrt{x}+1$

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

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

$=3x-3\sqrt{x}-2+x-1$

$=4x-3\sqrt{x}-3$

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

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

$=x-9-2x+3\sqrt{x}+2$

$=-x+3\sqrt{x}-7$

$E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)$

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

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

$=x-4-4x-6\sqrt{x}+4$

$=-3-6\sqrt{x}$

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QE

Quynh Existn't

17 tháng 7 2021

Rút gọn các biểu thức...

thực hiện phép tính$\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}$giải phương trình$\sqrt{x-3}=6$$\sqrt{\left(x-3\right)^2}=12$rút gọn biểu thứca) $P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right).\left(\dfrac{3-\sqrt{x}}{3-x}\right)$ (với x≥0 ; x≠3; x≠9b) $P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right)\div\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}$ (x >0)c) $A=\sqrt{3x-1}+3.\sqrt{12x-4}-\sqrt{6^2.\left(3x-1\right)}+\sqrt{5}$ với...

7

AH

Akai Haruma

Giáo viên

17 tháng 7 2021

1. ĐKXĐ: $x>0; x\neq 9$

$A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}$

Đúng(1)

AH

Akai Haruma

Giáo viên

17 tháng 7 2021

2. ĐKXĐ: $x\geq 0; x\neq 4$

$B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)$

$=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2$

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N

nini

2 tháng 11 2023

1

NL

Nguyễn Lê Phước Thịnh

9 tháng 11 2023

Bài 1:

$\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}$

$=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}$

$=4-\sqrt{5}+\sqrt{5}+1=5$

Bài 2:

a: ĐKXĐ: x>=3

$\sqrt{x-3}=6$

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: $x\in R$

$\sqrt{\left(x-3\right)^2}=12$

=>$\left|x-3\right|=12$

=>$\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.$

Bài 3:

a: $P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)$

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

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

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

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

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

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

c: $A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}$

$=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}$

$=\sqrt{3x-1}+\sqrt{5}$

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

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

$=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}$

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