Khi x = 4; y = 11 thì biểu thức đại số:              

\(\frac{y\left(x^2-1\right)}{xy+y}\) =  \(\frac{11\left(4^2-1\right)}{4.11+11}\)

= \(\frac{11.15}{11.5}\)

= 3

\(\frac{11\left(4^2-1\right)}{4\cdot11+4}=\frac{55}{16}\)

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

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

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

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

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

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

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 

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

      \(B=\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left[\left(x-y\right)\left(x+y\right)\right]^2}-\frac{x^2}{\left(x-y\right)\left(x+y\right)}\)  (làm tắt đấy x^2/(y^2 - x^2) = - x^2 /(x^2 - y^2)

Thay x + y = 1 vào B ta có 

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

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

A = \(\frac{y-x}{xy}:B=\frac{y-x}{xy}\cdot\frac{\left(x-y\right)^2}{\left(y^2-x^2y-x^3\right)}=\frac{\left(x-y\right)^3}{-xy\left(y^2-x^2y-x^3\right)}\)

Sorry mình không giúp đc bạn

a. \(=4x^2-4xy+y^2+4x^2-4xy+y^2=8x^2+2y^2\)

\(=8.\left(\frac{1}{21}\right)^2+4.\left(-0.3\right)^2=\frac{4169}{11025}\)

b, \(=\left(\frac{1}{7}xy+7yz+\frac{1}{7}xy-7yz\right)\left(\frac{1}{7}xy+7yz-\frac{1}{7}xy+7yz\right)\)

\(=\frac{2}{7}xy.14yz=4xy^2z=4.2.\left(0,25\right)^2.\left(-4\right)=-2\)