Ta có : \(x^2-2y^2=xy\)

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

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

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

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

Mà x+y khác 0 nên \(x-2y=0=>x=2y\)

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

\(ĐKXĐ:x\ne1\)

a) \(Q=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}+\frac{3-\sqrt{x}}{x-1}\)

\(=\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)+\sqrt{x}-3}{1-x}\)

\(=\frac{\sqrt{x}+x+\sqrt{x}-x+\sqrt{x}-3}{1-x}\)

\(=\frac{3\sqrt{x}-3}{1-x}=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-3}{\sqrt{x}+1}\)

b) Ta có \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}+1\ge1\)

\(\Rightarrow\frac{-3}{\sqrt{x}+1}\ge-3\)

Dấu "=" khi x = 0

a/ \(Q=\sqrt{x}+\sqrt{y}\)

b/ \(\hept{\begin{cases}x+y=2015\\xy=2016\end{cases}}\)

\(Q^2=x+y+2\sqrt{xy}=2015+2\sqrt{2016}\)

\(\Rightarrow Q=\sqrt{2015+2\sqrt{2016}}\)

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

Q\(=\left(x^2-xy+y^2-xy\right):\left[\sqrt{x}\left(x-y\right)-\sqrt{y}\left(x-y\right)\right]\)

Q\(=\left(x^2-2xy+y^2\right):\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\)

Q \(=\left(x-y\right)^2:\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\)

Q \(=\left(x-y\right):\left(\sqrt{x}-\sqrt{y}\right)\)

Q \(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right):\left(\sqrt{x}-\sqrt{y}\right)\)

Q \(=\sqrt{x}+\sqrt{y}\)

\(A=\frac{3}{2+\sqrt{-x^2+2x+7}}=\frac{3}{2+\sqrt{8-\left(x^2-2x+1\right)}}=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\)

Vì \(\left(x-1\right)^2\le0\Rightarrow8-\left(x-1\right)^2\le8\Rightarrow\sqrt{8-\left(x-1\right)^2}\le\sqrt{8}\)

\(\Rightarrow2+\sqrt{8-\left(x-1\right)^2}\le2+\sqrt{8}\)=>\(A=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\ge\frac{3}{2+\sqrt{8}}\)

Dấu "=" xảy ra khi x=1

Vậy minA=\(\frac{3}{2+\sqrt{8}}\) khi x=1

co mik

t chơi