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\(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)( ĐKXĐ tự tìm nhé *)
\(=\frac{\left(x^3\right)^2+2x^3y^3+\left(y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\frac{\left(x^3+y^3\right)^2}{x\left[\left(x^3\right)^2-\left(y^3\right)^2\right]}\)
\(=\frac{\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}\)
\(=\frac{\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^3+y^3}{x\left(x^3-y^3\right)}=\frac{x^3+y^3}{x^4-xy^3}\)
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\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
Ta thay : \(x=0;y=2009;z=2010\) ta được :
\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)
Chúc bạn học tốt !!!
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)
Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
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Bạn viết rõ hơn nhé :
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
= \(\frac{x^4-xy^3}{2xy+y^2}.\frac{2x+y}{x^3+x^2y+xy^2}\)
= \(\frac{x.\left(x-y\right).\left(x^2+xy+y^2\right).\left(2x+y\right)}{y.\left(2x+y\right).x.\left(x^2+xy+y^2\right)}\)
= \(\frac{x-y}{y}\)
Chúc bạn học tốt !!!
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a, \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)=\left(x+2\right)^2=x^2+4x+4\)
b, \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-27x\right)=x^3-27-x^2+27x\)
c, \(\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3+y^3-x^3+y^3=2y^3\)
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\(\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+y^3+3x^2y+3xy^2-x^3+y^3+3x^2y-3xy^2-2y^3\)
\(=2y^3+6x^2y-2y^3\)
\(=6x^2y\)
Với a+b+c=0 thì \(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3=a^3+b^3-a^3-b^3-3ab\left(a+b\right)\)
\(=-3ab\left(a+b\right)=-3ab.\left(-c\right)=3abc\)
Áp dụng:\(\left(x+y\right)^3-\left(x-y\right)^3-\left(2y\right)^3=\left(x+y\right)^3+\left(y-x\right)+\left(-2y\right)^3\)
có \(x+y+y-x+\left(-2y\right)=0\)\(\Rightarrow\left(x+y\right)^3-\left(x-y\right)^3-\left(2y\right)^3=3.\left(-2y\right).\left(x+y\right)\left(y-x\right)\)
\(=6y\left(x+y\right)\left(x-y\right)\)\(=6y\left(x^2-y^2\right)=6x^2y-6y^3\)
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a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
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a)<=>
A,=(x+y)(x-y)=x^2-y^2
x=(-1/2)^5:(1/2)^4=-1/2
x^2=1/4
y=8^2/(-2)^5=-2
y^2=4
A=1/4-4=-15/4
\(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\)
ĐKXĐ : \(x+y\ne0\Leftrightarrow x\ne-y\)
\(=\frac{5\cdot3x\cdot\left(x+y\right)^2\left(x+y\right)}{5\cdot y\cdot\left(x+y\right)^2}\)
\(=\frac{3x\left(x+y\right)}{y}\)