\(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}=x+2y-\left|x-2y\right|=x+2y-x+2y=4y\)

\(x-2y-\sqrt{x^2-4xy+4y^2}\left(1\right)=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)

TH1: \(x\ge2y\)

\(\left(1\right)=x-2y-x+2y=0\)

TH2: \(x< 2y\)

\(\left(1\right)=x-2y+x-2y=2x-4y\)

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

= x - 2y - /x-2y/

= x - 2y - x + 2y

= 0

\(x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\left|x-2y\right|\)

\(=\left[{}\begin{matrix}x-2y-x+2y=0\left(x\ge2y\right)\\x-2y+x-2y=2x-4y\left(x< 2y\right)\end{matrix}\right.\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)

Lời giải:

$x-2y-\sqrt{x^2-4xy+4y^2}$

$=x-2y-\sqrt{(x-2y)^2}=x-2y-|x-2y|$

Nếu $x\geq 2y$ thì $|x-2y|=x-2y$

Khi đó: $x-2y-\sqrt{x^2-4xy+4y^2}=0$

Nếu $x< 2y$ thì $|x-2y|=2y-x$

Khi đó: $x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-(2y-x)=2(x-2y)$

\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)

\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)

\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)

a: \(=\dfrac{\left|x+2\right|}{x-1}\)

b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)

c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)

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

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

A = x - 2y - x + 2y

A = 0

a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)

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

\(=x-2y-\left|x-2y\right|\)

TH1: \(x-2y--\left(x-2y\right)\)

\(=x-2y+x-2y\)

\(=2x-4y\)

TH2: \(x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

b) \(x^2+\sqrt{x^4-8x^2+16}\)

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

\(=x^2+\left|x^2-4\right|\)

TH1: 

\(x^2+-\left(x^2-4\right)\)

\(=x^2-x^2+4\)

\(=4\)

TH2: 

\(x^2+\left(x^2-4\right)\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)

\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)

\(=2x-1-\sqrt{x-5}\)

d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))

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

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

e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)

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

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

\(=\left|x^2-4\right|+1\)

TH1: 

\(x^2-4+1\)

\(=x^2-3\)

TH2:

\(-\left(x^2-4\right)+1\)

\(=-x^2+4+1\)

\(=-x^2+5\)

a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)

=x-2y-|x-2y|

Khi x>=2y thì A=x-2y-x+2y=0

Khi x<2y thì A=x-2y+x-2y=2x-4y

b: \(B=x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\left|x^2-4\right|\)

TH1: x>=2 hoặc x<=-2

B=x^2+x^2-4=2x^2-4

TH2: -2<=x<=2

B=x^2+4-x^2=4

c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)

\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)

d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)

Lời giải:

Sửa đề: Rút gọn \(x+2y-\sqrt{x^2-4xy+4y^2}\) \((x\geq 2y)\)

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Ta có:

\(x+2y-\sqrt{x^2-4xy+4y^2}=x+2y-\sqrt{x^2-2.x.2y+(2y)^2}\)

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

\(=x+2y-|x-2y|=x+2y-(x-2y)=4y\)

(do \(x\geq 2y\Rightarrow |x-2y|=x-2y\) )