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Chúc bạn học tốt

\(x^2-2y^2=xy\Rightarrow x^2-2y^2-xy=0\Rightarrow x^2-y^2-y^2-xy=0\)

\(\Rightarrow\left(x+y\right)\left(x-y\right)-y\left(x+y\right)=0\)

\(\Rightarrow\left(x+y\right)\left(x-2y\right)=0\Rightarrow x-2y=0\)\(\left(x+y\ne0\right)\)

\(\Rightarrow x=2y\)

Thay vào A tính đc giá trị của A

Ta có \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2y\\x=-y\end{cases}}\)

với x=2y, thao vào, ta có A=1/3

với x=-y thay vào không thỏa mãn 

^.^

\(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)

\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)

\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\) 

\(\Rightarrow x-2y=0\) vì \(x+y\ne0\)

\(\Leftrightarrow x=2y\Rightarrow A=\frac{2y-y}{2y+y}=\frac{1}{3}\)

Ta có: \(x-y-xy=0\)

\(\Leftrightarrow x-y.\left(1+x\right)=0\)

\(\Leftrightarrow\left(1+x\right)-y.\left(1+x\right)=0+1\)

\(\Leftrightarrow\left(1+x\right).\left(1-y\right)=1\)

Bạn tìm x,y rùi tính \(\frac{1}{x}-\frac{1}{y}\)nhé

giup minh voi

\(a)\) \(\frac{x^2y-xy}{x-1}=xy\)

\(\Leftrightarrow\)\(\frac{xy\left(x-1\right)}{x-1}=xy\)

\(\Leftrightarrow\)\(xy=xy\) ( đpcm ) 

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

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

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

\(\Leftrightarrow\)\(x\left(x+y\right)=x^2+xy^2\)

\(\Leftrightarrow\)\(x^2+xy=x^2+xy^2\)

\(\Leftrightarrow\)\(xy=xy^2\)

\(\Leftrightarrow\)\(y=y^2\) ( đề sai hay mình sai =.= ) 

Chúc bạn học tốt ~ 

a, \(\frac{x^2y-xy}{x-1}=\frac{xy\left(x-1\right)}{x-1}=xy\)

b,Sửa đề \(\frac{x^2-y^2}{x^2+xy}=\frac{x-y}{x}\)

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

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

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

\(=4-\dfrac{3\left(x-y\right)^2}{x^2+y^2-xy}\le4\)

\(P_{max}=4\) khi \(x=y=\dfrac{1}{2}\)

\(x^2-2y^2=xy\)

\(\Leftrightarrow\left(x^2+xy\right)-\left(2y^2-2xy\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

\(\Rightarrow x=2y\)

\(\Rightarrow A=\frac{x-y}{x+y}=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)