Bài làm:

Ta có:

\(\frac{\sqrt{50x^4y^2}}{\sqrt{200x^2}y^4}=\frac{5x^2y\sqrt{2}}{10xy^4\sqrt{2}}=\frac{x}{2y^3}\)

a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)

=x

b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)

\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)

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

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

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

=4x

Ta có: \(x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}\)

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

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

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

Thay vào ta tính được:

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

Mà \(x^2+y^2-\frac{x^2y^2}{x^2+y^2}=\frac{\left(x^2+y^2\right)^2-x^2y^2}{x^2+y^2}=\frac{x^4+x^2y^2+y^4}{x^2+y^2}>0\left(\forall x,y\right)\)

Khi đó:

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

\(P=\sqrt{x^2+y^2+\frac{x^2y^2}{\left(x+y\right)^2}}\)

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

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

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

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

Vậy \(P=\frac{x^2+xy+y^2}{x+y}\)

a)\(B=\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{0}+\sqrt{4}}=\frac{1}{2}\)

b)\(M=A+B=\frac{2\sqrt{y}}{x-y}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)\(=\frac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)

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

c)\(M=\frac{2}{\sqrt{x}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{4y}-\sqrt{y}}\)<=>\(1=\frac{2}{2\sqrt{y}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{y}}\)<=> \(\sqrt{y}=2\)

<=> \(\left(\sqrt{y}\right)^2=2^2\)<=> \(y=4\)

=>\(x=4y=4\cdot4=16\)

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

 k mình đi mình sẽ giúp, mình rất cần người như cậu  Ngân 

Bạn phải giúp thì mình mới k chứ nhỉ ??

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

\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\left|\frac{x^2}{y^2}\right|=-5\)

\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\)

\(0.2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0.2.4\sqrt{\frac{x^6y^6}{x^4y^8}}=0.8\frac{x}{y}\)