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8 tháng 6 2018

\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

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

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

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

AH
Akai Haruma
Giáo viên
22 tháng 6 2023

1.

$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$

$=x+3+(3-x)=6$

2.

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

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

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

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

 

AH
Akai Haruma
Giáo viên
22 tháng 6 2023

4.

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

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

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$

1 tháng 7 2021

ĐKXĐ: \(x\ge2\)

\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)

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

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

Xét \(x\ge4\Rightarrow\sqrt{x-2}\ge\sqrt{2}\)

\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)

Xét \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)

\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)

1 tháng 7 2021

Tại sao xét  x≥4 vậy bạn.

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

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

TH1: x>=1

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

TH2: 1/2<=x<1

\(A=\dfrac{1}{\sqrt{2}}\left(1-\sqrt{2x-1}-\sqrt{2x-1}-1\right)=-\sqrt{4x-2}\)

2: 

\(=\sqrt{x-1+6\sqrt{x-1}+9}-\sqrt{x-2-2\sqrt{x-2}+1+3}\)

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

23 tháng 9 2021

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{5x+10\sqrt{x}-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5x+10\sqrt{x}-5\sqrt{x}+6+x-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

23 tháng 9 2021

\(P=\dfrac{5\sqrt{x}}{\sqrt{x}-2}-\dfrac{3-\sqrt{x}}{\sqrt{x}+2}+\dfrac{6x}{4-x}\left(đk:x\ge0,x\ne4\right)\)

\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+2\right)-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{5x+10\sqrt{x}+x-5\sqrt{x}+6-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{5\sqrt{x}+6}{x-4}\)

14 tháng 12 2021

\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

14 tháng 12 2021

Cảm ơn nhiều

Ta có: \(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}+2}\)

9 tháng 9 2021

\(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\\ B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)

\(P=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)

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

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)

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

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

\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

24 tháng 9 2021

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)