Ta có : \(B=\frac{y}{2}+\frac{3}{4}\sqrt{1-4y+4y^2}-\frac{3}{2}\)

=> \(B=\frac{2y-6}{4}+\frac{3}{4}\sqrt{\left(2y-1\right)^2}\)

=> \(B=\frac{\left(2y-6+3\sqrt{\left(2y-1\right)^2}\right)}{4}\)

=> \(B=\frac{\left(2y-6+3\left(1-2y\right)\right)}{4}\)

=> \(B=\frac{\left(2y-6+3-6y\right)}{4}=\frac{-4y-3}{4}=-\left(y+\frac{3}{4}\right)\)

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)

\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)

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

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

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

Giờ chỉ cần thế x, y vô nữa là xong nhé.

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)

\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)

\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)

\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)

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

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

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

\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

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

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

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

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

Ừ nhở chị sai từ chỗ \(\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2}{y\left(x+y\right)}=\frac{y^2+2xy}{y\left(x+y\right)}\)em nhé 

a/ \(\frac{7x-14y}{x^2-4y^2}=\frac{7\left(x-2y\right)}{x^2-\left(2y\right)^2}=\frac{7\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{7}{x+2y}.\)

b/ \(\frac{1-\frac{2y}{x}+\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x^2-2xy+y^2}{x^2}}{\frac{y-x}{xy}}=\frac{\left(x-y\right)^2}{x^2}.\frac{xy}{-\left(x-y\right)}=-\frac{y\left(x-y\right)}{x}=\frac{y\left(y-x\right)}{x}\)

1,

\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)

\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)

\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)

\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

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

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

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)

a)\(N=\left(\frac{x^2}{x^2-y^2}+\frac{y}{x-y}\right):\frac{x^3-y^3}{x^5-x^4y-xy^4+y^5}\)

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

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

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

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

b) Ta có: \(x+y=\frac{1}{40}\)

\(\Rightarrow\left(x+y\right)^2=\frac{1}{1600}\)

\(\Rightarrow x^2+2xy+y^2=\frac{1}{1600}\)

\(\Rightarrow x^2-\frac{1}{40}+y^2=\frac{1}{1600}\)

\(\Rightarrow x^2+y^2=\frac{1}{1600}+\frac{1}{40}\)

\(\Rightarrow x^2+y^2=\frac{41}{1600}\)

Vậy \(N=\frac{41}{1600}\)

\(=\frac{\sqrt{y}}{\sqrt{y}-2}\times\frac{\left(\sqrt{y}-2\right)\left(\sqrt{y}+2\right)}{\sqrt{4}\cdot\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}+2}\times\frac{\left(\sqrt{y}+2\right)\left(\sqrt{y}-2\right)}{\sqrt{4}\cdot\sqrt{y}}\)

\(=\frac{\sqrt{y}+2}{\sqrt{4}}+\frac{\sqrt{y}-2}{\sqrt{4}}=\frac{2\sqrt{y}}{2}=\sqrt{y}\)

b/ đkxd \(y>0;y\ne4\)

tại  \(y=\frac{1}{4}\)( t/m dkxd )  nên \(P=\sqrt{y}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

a/ \(\frac{y}{x}.\left(\sqrt{\frac{x^2}{y^4}}\right)=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)

b/ \(2y^2.\sqrt{\frac{x^4}{4y^2}}=2y^2.\sqrt{\frac{\left(x^2\right)^2}{\left(-2y\right)^2}}=2y^2.\frac{x^2}{-2y}=-y.x^2\)

c/ \(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\sqrt{\frac{\left(-5x\right)^2}{\left(y^3\right)^2}}=5xy.\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)

d/\(0,2.x^3y^3.\sqrt{\frac{4^2}{\left(x^2y^4\right)^2}}=\frac{1}{5}.x^3y^3.\frac{4}{x^2y^4}=\frac{4x}{5y}\)

Trần Việt Linh sai phần b,c,d r bn

Sửa lại:

b) 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\) với y<0

Ta có : 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\)=2y\(^2\).\(\frac{x^2}{\left|y\right|}\)

Vì y>0 nên |y| = -y.Ta có : 2y\(^2\).\(\frac{x^2}{2\left|y\right|}\)= -2y\(^2\).\(\frac{x^2}{2y}\) = -2x\(^2\)y

c) 5xy.\(\sqrt{\frac{25x^2}{y^6}}\) với x<0,y>0

Ta có :5xy\(\sqrt{\frac{25x^2}{y^6}}\)=5xy.\(\frac{5\left|x\right|}{y^3}\) ( y>0)

Vì x<0 nên |x| =-x .Ta có : 5xy.\(\frac{5\left|x\right|}{y^3}\)= -5xy.\(\frac{5x}{y^3}\) =\(\frac{-25x^2}{y^2}\)

d) 0,,2x\(^3\)y\(^3\).\(\sqrt{\frac{16}{x^4y^8}}\) với x#o,y#0

Ta có: 0,2x\(^3\)y\(^3\)\(\frac{4}{x^2y^4}\)=\(\frac{0,8x}{y}\) ( vì #0,y#0)

a)\(\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)

b) \(2y^2\cdot\sqrt{\frac{x^4}{4y^2}}=2y^2\cdot\frac{x^2}{2y}=x^2y\)

c) \(5xy\cdot\sqrt{\frac{25x^2}{4y^6}}=5xy\cdot\frac{5x}{2y^3}=\frac{5x^2y}{2y^3}=\frac{5x^2}{2y^2}\)

d) \(2x^3y^3\cdot\sqrt{\frac{16}{x^4y^8}}=2x^3y^3\cdot\frac{4}{x^2y^4}=\frac{8x}{y}\)

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

b/ Khi \(x=4y\) và M=1

\(\Leftrightarrow\frac{2}{\sqrt{4y}-\sqrt{y}}=1\Leftrightarrow\frac{2}{2\sqrt{y}-\sqrt{y}}=1\Leftrightarrow\frac{2}{\sqrt{y}}=1\)

\(\Leftrightarrow\sqrt{y}=2\Rightarrow y=4\Rightarrow x=16\)