\(\frac{5x^2-10xy+5y^2}{2x^2-2xy+2^2}:\frac{8y-8x}{x^2-xy+y^2}\)
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NP
1
VM
1 tháng 2 2020
Bài 4:
a) \(\frac{2x^2-10xy}{2xy}+\frac{5y-x}{y}\)
\(=\frac{y.\left(2x^2-10xy\right)}{2xy.y}+\frac{2xy.\left(5y-x\right)}{2xy.y}\)
\(=\frac{2x^2y-10xy^2}{2xy^2}+\frac{10xy^2-2x^2y}{2xy^2}\)
\(=\frac{2x^2y-10xy^2+10xy^2-2x^2y}{2xy^2}\)
\(=\frac{0}{2xy^2}\)
\(=0.\)
b) \(\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{x^2-y^2}\)
\(=\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{1.\left(x+y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y}{\left(x-y\right).\left(x+y\right)}+\frac{x+y}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y+x+y+3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x-y}{\left(x-y\right).\left(x+y\right)}\)
c) \(x+y+\frac{x^2+y^2}{x+y}\)
\(=\frac{x+y}{1}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right).\left(x+y\right)}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right)^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2+x^2+y^2}{x+y}\)
\(=\frac{2x^2+2xy+2y^2}{x+y}.\)
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QT
0
LB
23 tháng 8 2017
P=\(X^2+2Y^2-2XY+8X+8Y+2017\)
P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)
P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)
DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)