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Ta có:
VT=(x2+y2)2−(2xy)2VT=(x2+y2)2−(2xy)2
=(x2+y2−2xy)(x2+y2+2xy)=(x2+y2−2xy)(x2+y2+2xy)
=(x−y)2(x+y)2=VP=(x−y)2(x+y)2=VP
⇒đpcm⇒đpcm
`x/(x+y) + (2xy)/(x^2-y^2) - y(x+y)`
`= (x(x-y))/(x^2-y^2) + (2xy)/(x^2-y^2) - (y(x-y))/(x^2-y^2)`
`= (x^2 - xy + 2xy - xy + y^2)/(x^2-y^2)`
`= (x^2+y^2)/(x^2-y^2)`
\(\dfrac{x}{x+y}+\dfrac{2xy}{x^2-y^2}-\dfrac{y}{x+y}\)
\(=\dfrac{x-y}{x+y}+\dfrac{2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x+y\right)\left(x-y\right)}+\dfrac{2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2+y^2}{x^2-y^2}\)
\(=\left[\frac{2xy}{\left(x-y\right).\left(x+y\right)}+\frac{x-y}{2.\left(x+y\right)}\right]:\frac{x+y}{2x}+\frac{x}{y-x}\)
\(=\frac{4xy+\left(x-y\right).\left(x-y\right)}{2.\left(x-y\right).\left(x+y\right)}.\frac{2x}{x+y}+\frac{x}{y-x}\)
\(=\frac{x^2+2xy+y^2}{\left(x-y\right).\left(x+y\right)^2}.x+\frac{x}{y-x}\)
\(=\frac{x.\left(x+y\right)^2}{\left(x-y\right).\left(x+y\right)^2}+\frac{x}{y-x}\)
\(=\frac{x}{x-y}-\frac{x}{x-y}=0\)
Bạn giùm mik nhé, tks bạn nhiều (:
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
Ta có: \(\dfrac{y}{x-y}-\dfrac{x^3-xy^2}{x^2+y^2}\cdot\left(\dfrac{x}{x^2-2xy+y^2}-\dfrac{y}{x^2-y^2}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\left(\dfrac{x\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}-\dfrac{y\cdot\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x\cdot\left(x^2+y^2\right)}{\left(x^2+y^2\right)\cdot\left(x-y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)
a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
`@` `\text {Ans}`
`\downarrow`
\(( x + y ) ( x^2 + 2xy + y^2 )\)
`= x(x^2 +2xy + y^2) + y(x^2 + 2xy + y^2)`
`= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3`
`= x^3 + 3x^2y + 3xy^2 + y^3`