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=\(\left(\frac{1}{x\left(x-y\right)}-\frac{3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(\left(\frac{y\left(x+y\right)+x^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy+y^2-3y^2-y\left(x-y\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\) \(\left(\frac{x^2+xy+y^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy-2y^2-xy+y^2}{x\left(x-y\right)}\right)\left(\frac{1}{x+y}\right)\)
=\(\frac{x^2-y^2}{x\left(x-y\right)\left(x+y\right)}\)=\(\frac{\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}\) =\(\frac{1}{x}\)
mk ko biết cứ bấm đại thui, bn có thể giúp mk ko ???
\(=\dfrac{a+x+1}{a+x}:\dfrac{a+x-1}{a+x}\cdot\left(\dfrac{2ax-1+a^2+x^2}{2ax}\right)\)
\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x\right)^2-1}{2ax}\)
\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x+1\right)\left(a+x-1\right)}{2ax}\)
\(=\dfrac{\left(a+x+1\right)^2}{2ax}\)
\(=\dfrac{x^3-1}{x}\cdot\dfrac{x^2-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2+x}{x}=2x+1\)
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
\(A=\frac{x^2}{x^2-1}-\frac{x^2}{x^2+1}\left(\frac{x}{x+1}+\frac{1}{x^2+x}\right)\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x}{x+1}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x^2}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}.\frac{x^2+1}{x\left(x+1\right)}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x}{x+1}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
=>\(A=\frac{x^2-x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x^2-x^2+x}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x}{x^2-1}\)
\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\left(ĐK:x\ne0;y\ne0\right)\)
\(=\frac{2}{xy}:\left(\frac{y-x}{xy}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}\cdot\frac{x^2y^2}{\left(y-x\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{-2xy}{\left(x-y\right)^2}+\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{-2xy+x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{\left(x-y\right)^2}{\left(x-y\right)^2}=1\)
\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\left(dk:x\ne y\ne0\right)\)
miik ko nghĩ nó là toán lớp 7 đâu bn