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

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

\(=\dfrac{x+2y}{\left(x-2y\right)\left(x+2y\right)}-\dfrac{x}{x+2}\cdot\dfrac{\left(x+2\right)^2}{x\left(x-2y\right)}=\dfrac{1}{x-2y}-\dfrac{x+2}{x-2y}=\dfrac{-x-1}{x-2y}\)

Sửa đề: \(A=\dfrac{1-3x}{2y}\cdot\sqrt{\dfrac{36y^2}{9x^2-6x+1}}\)

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

\(=\dfrac{1-3x}{2y}\cdot\left|\dfrac{6y}{3x-1}\right|\)

x>1/3 nên 3x-1>0

y>0 nên 6y>0

=>\(A=\dfrac{1-3x}{2y}\cdot\dfrac{6y}{3x-1}=-3y\)

a) = . = . = vì x > 0.

Do đó = .

b) = . = ..

Vì y < 0 nên │y│= -y. Do đó = . = .

c) 5xy. = 5xy. = 5xy..

Vì x < 0, y > 0 nên = -x và = .

Do đó: 5xy = 5xy. = -.

d) 0,2 = = 0,2 =

Nếu x > 0 thì > 0 nên . Do đó 0,2 = .

Nếu x < 0 thì < 0 nên . Do đó 0,2 = -.

\(a,=\dfrac{x}{y}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{x^2}{y^3}\\ b,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\)

(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)

(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)

(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)

(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)

a) 1/y 

b) - x^2 y 

c) -25x^2 / y^2

d) 4x/5y

yx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.

(Do $x>0$ nên $|x|=x$, $y\ne 0$ $\Rightarrow$ $y^2>0$ nên $|y^2|=y^2$)

x44y2=2y2.x44y2=2y2.|x2||2y|=2y2.x2−2y=2y2.x2−2y=−x2y.

(Do $y<0$ nên $|2y|=-2y$ và $x^2\ge 0$ nên $|x^2|=x^2$)

25x2y6=5xy.25x2y6=5xy.|5x||y3|=5xy.−5xy3=5xy.(−5x)y3=−25x2y2.

(Do $x<0$ nên $|5x|=-5x$ và $y>0$ $\Rightarrow$ $y^3>0$ nên $|y^3|=y^3$)

16x4y8=0,2x3y3.16x4y8=210.x3y3.4|x2y4|=15.x3y3.4x2y4=4x3y35x2y4=4x5y.

( Do $x\ne 0$ $\Rightarrow$ $x^2>0$ và $y\ne 0$ $\Rightarrow$ $y^4>0$ nên $|x^2{y}^4|=x^2{y}^4$)

a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)

\(=-4.\dfrac{1}{4}+10=9\)

b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)

\(=\left(-2\right).\left(32-32\right)=0\)

a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)

\(=4x^2-4x+1+9-4x^2\)

\(=-4x+10\)

\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)

đề bài sai thì phải

bạn Hoàng Thanh Tuấn nói đúng đấy

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

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

\(\Leftrightarrow B=x+2y-\left(x^2-2y\right)\)

\(\Leftrightarrow B=x+2y-x^2+2y\)

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

Đến đây mình không biết giải thế nào nữa

Nếu đề sai thì mình nghĩ là: \(B=x^2+2y-\sqrt{x^4-4xy+4y^2}=4y\)(Làm như trên)

a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\) 

b) \(B=\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\) 

\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)

\(B=\left[\dfrac{\left(y-x\right)\left(x+y\right)}{\left(x-2y\right)\left(x+y\right)}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

\(B=\dfrac{y^2-x^2-x^2-y^2-y+2}{\left(x+y\right)\left(x-2y\right)}:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

\(B=\dfrac{-2x^2-y+2}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)

\(B=\dfrac{-\left(2x^2+y-2\right)}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)

\(B=\dfrac{-\left(x+1\right)}{\left(x-2y\right)\left(2x^2+y+2\right)}\)