rút gọn:
x√x +y√y+x-y
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\(A=x^2-2xy+y^2+x^2+2xy+y^2+x^2-y^2=3x^2+y^2\\ B=\left(x-y-x+y-z\right)^2=\left(-z\right)^2=z^2\)
\(2\left(x-y\right)\left(x++y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
\(=4x^2\)
Lời giải:
a) ĐK: $x\geq 0; y\geq 0; x\neq y$
\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
Lời giải:
$x(x+y)-y(x+y)+x^2+y^2=(x-y)(x+y)+x^2+y^2$
$=x^2-y^2+x^2+y^2=2x^2$
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
\(A=\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}+\dfrac{y}{y-x}\left(ĐKXĐ:x\ne\pm y\right)\)
\(A=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x+y\right)\left(x-y\right)}\right):\dfrac{x+y}{2x}+\dfrac{y}{y-x}\)
\(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}+\dfrac{y}{y-x}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}+\dfrac{y}{y-x}\)
\(\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}+\dfrac{y}{y-x}=\dfrac{x}{x-y}+\dfrac{y}{y-x}=\dfrac{x}{x-y}-\dfrac{y}{x-y}=\dfrac{x-y}{x-y}=1\)
x(x – y) + y(x – y)
= x.x – x.y + y.x – y.y
= x2 – xy + xy – y2
= x2 – y2 + (xy – xy)
= x2 – y2
\(x\sqrt{x}+y\sqrt{y}+x-y\)
\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}\right)\)
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