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=\(\left(\frac{1}{x\left(x-y\right)}-\frac{3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(\left(\frac{y\left(x+y\right)+x^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy+y^2-3y^2-y\left(x-y\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\) \(\left(\frac{x^2+xy+y^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy-2y^2-xy+y^2}{x\left(x-y\right)}\right)\left(\frac{1}{x+y}\right)\)
=\(\frac{x^2-y^2}{x\left(x-y\right)\left(x+y\right)}\)=\(\frac{\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}\) =\(\frac{1}{x}\)
\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\left(ĐK:x\ne0;y\ne0\right)\)
\(=\frac{2}{xy}:\left(\frac{y-x}{xy}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}\cdot\frac{x^2y^2}{\left(y-x\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{-2xy}{\left(x-y\right)^2}+\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{-2xy+x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{\left(x-y\right)^2}{\left(x-y\right)^2}=1\)
\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\left(dk:x\ne y\ne0\right)\)
miik ko nghĩ nó là toán lớp 7 đâu bn
=\(\frac{36-4+3}{6}-\frac{30+10-9}{6}-\frac{18-14+15}{6}\)
=\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}-\frac{15}{6}=-\frac{5}{2}\)
bài này thì dễ:
\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
Cách 1:
\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}=-\frac{15}{6}=-\frac{5}{2}\)
Cách 2:
\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}=-2\frac{1}{2}=-\frac{5}{2}\)
mk ko biết cứ bấm đại thui, bn có thể giúp mk ko ???
A=\(\left[\frac{x\left(x-y\right)}{y\left(x+y\right)}+\frac{\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\right]:\left[\frac{y^2}{x\left(x-y\right)\left(x+y\right)}+\frac{1}{x+y}\right]\frac{ }{ }\)
=\(\left[\frac{x^2\left(x-y\right)+y\left(x-y\right)\left(x+y\right)}{xy\left(x+y\right)}\right]:\left[\frac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\right]\)=\(\frac{\left(x-y\right)\left(x^2+y^2+xy\right)}{xy\left(x+y\right)}.\frac{x\left(x-y\right)\left(x+y\right)}{y^2+x\left(x-y\right)}\)
=\(\frac{\left(x-y\right)^2\left(x^2+y^2+xy\right)}{y\left(x^2+y^2-xy\right)}\)=\(\frac{\left(x-y\right)^2\left(x^2+xy+\frac{y^2}{4}+\frac{3y^2}{4}\right)}{y\left(x^2-xy+\frac{y^2}{4}+\frac{3y^2}{4}\right)}\)=\(\frac{\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{y.\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}\)
Ta nhận thấy các số trong ngoặc đều dương.
=> Để A>0 thì y>0
Vậy để A>0 thì y>0 và với mọi x
\(B=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^2}{y}-\frac{y^2}{x}\right)\right]:\frac{x-y}{x}\)
=>\(B=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^3}{xy}-\frac{y^3}{xy}\right)\right].\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{1}{x+y}.\frac{x^3-y^3}{xy}\right).\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{1}{x+y}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy}\right).\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{x^2-xy+y^2}{xy}\right).\frac{x}{x-y}\)
=>\(B=\frac{x^2-y^2-x^2+xy-y^2}{xy}.\frac{x}{x-y}\)
=>\(B=\frac{xy}{xy}.\frac{x}{x-y}\)
=>\(B=1.\frac{x}{x-y}\)
=>\(B=\frac{x}{x-y}\)