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MT
13 tháng 8 2015
\(\left(\frac{1}{x^2-xy}-\frac{3y^2}{x^4-xy^3}-\frac{9}{x^3+x^2y+xy^2}\right).\left(y+\frac{x^2}{x+y}\right)\)
\(=\left(\frac{1}{x.\left(x-y\right)}-\frac{3y^2}{x.\left(x^3-y^3\right)}-\frac{9}{x.\left(x^2+xy+y^2\right)}\right).\left(\frac{y.\left(x+y\right)}{x+y}+\frac{x^2}{x+y}\right)\)
\(=\left(\frac{1}{x.\left(x-y\right)}-\frac{3y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{9}{x.\left(x^2+xy+y^2\right)}\right).\left(\frac{y^2+xy}{x+y}+\frac{x^2}{x+y}\right)\)
\(=\left(\frac{x^2+xy+y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{9x-9y}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}\right).\left(\frac{x^2+xy+y^2}{x+y}\right)\)
\(=\frac{x^2+xy-2y^2-9x+9y}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}.\frac{x^2+xy+y^2}{x+y}\)
làm tip nha bận rồi
17 tháng 10 2019
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
17 tháng 10 2019
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
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 \(x=\frac{1}{2};y=\frac{1}{3}\)
BT
24 tháng 3 2018
\(A=\left[\frac{x^2-y^2}{xy}-\frac{1}{xy}\left(\frac{x^2}{y}-\frac{y^2}{x}\right)\right]:\frac{x-y}{xy}\)
\(A=\left[\frac{x^2-y^2}{xy}-\left(\frac{x}{y^2}-\frac{y}{x^2}\right)\right].\frac{xy}{x-y}\) => \(A=\left(\frac{x^2-y^2}{xy}-\frac{x^3-y^3}{x^2y^2}\right).\frac{xy}{x-y}=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^2y^2}\right).\frac{xy}{x-y}\)
=> \(A=\frac{x-y}{xy}\left(\left(x+y\right)-\frac{x^2+xy+y^2}{xy}\right).\frac{xy}{x-y}\)=> \(A=x+y-\frac{x^2+xy+y^2}{xy}=\frac{x^2y+xy^2-x^2-xy-y^2}{xy}\)
7 tháng 11 2018
a)\(\frac{3xy+6}{6xy+12}=\frac{1}{2}\Leftrightarrow\left(3xy+6\right)\cdot2=\left(6xy+12\right)\cdot1\)
\(\Leftrightarrow6xy+12=6xy+12\)
Vậy.......
b)\(\frac{x^2-xy}{5y^2-5xy}=\frac{x}{5y}\Leftrightarrow\left(x^2-xy\right)\cdot5y=\left(5y^2-5xy\right)\cdot x\)
\(\Leftrightarrow5x^2y-5xy^2=5xy^2-5x^2y\)
Vậy.....
\(\frac{x^5y}{\left(xy\right)^4}\)
\(=\frac{x^5y}{x^4y^4}=\frac{x}{y^3}\)