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5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y^2\right)\)
\(=4y^2\)
6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)
\(=\left(-3\right)^2=9\)
8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)
\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)
\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)
\(=\left(4x\right)^2=16x^2\)
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
\(\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)2xy}\)
=\(\frac{x^2+5x+y^2+5y+2xy-3}{x^2+6x+y^2+6y+2xy}\)
triệt tiêu x2;y2;2xy ta được:
\(\frac{5x+5y-3}{6x+6y}=\frac{5\left(x+y\right)-3}{6\left(x+y\right)}\)
=\(\frac{5.2010-3}{6.2010}=\frac{3349}{4020}\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
\(\frac{\left(x+y\right)^7-x^7-y^7}{\left(x+y\right)^5-x^5-y^5}=\left(x+y\right)^2-x^2-y^2\)
=x2+2xy+y2-x2-y2
=x2-x2+y2-y2+2xy
= 2xy