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1 tháng 2 2020
a/VT=x5+x^4.y+x^3.y^2+x^2.y^4+x.y^4-x^4.y-x^3.y^2-x^2.y^3-x.y^4-y^5
=x^5-y^5=VP
=>dpcm
17 tháng 7 2015
Nhiều quá
a, ( x+ y) = - p => ( x + y)^2 = p^2
=> x^2 + 2xy + y^2 = p^2
=> x^2 + 2q + y^2 =p^2
=> x^2 + y^2 = p^2 - 2q
S
27 tháng 7 2019
\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)
\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)
S
27 tháng 7 2019
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)
TT
0
NC
14 tháng 8 2020
a) \(\left(x+y-z\right)^2=\left[\left(x+y\right)-z\right]^2\)
\(=\left(x+y\right)^2-2\left(x+y\right)z+z^2\)
\(=x^2+2xy+y^2-2zx-2yz+z^2\)
\(=x^2+y^2+z^2+2xy-2yz-2zx\)
b) \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4\)
c) \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5\)
LO
1
22 tháng 7 2017
a, \(x^3+y^3+z^3=3xyz\Rightarrow x^3+y^3+z^3-3xyz=0\)( 1 )
Nhận xét : \(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\Rightarrow x^3+y^3=\left(x+y\right)^3-3x^2-3xy^2\)
Thay vào ( 1 ) ta có :
\(\left(x+y\right)^3+c^3-3x^2y-3xy^2-3xyz\)
\(=\left(z+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(z+y+z\right)\left(z^2+2xy+y^2-xz-yz+z^2\right)-3xyz\left(z+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(z^2+x^2+y^2-xy-yz-xz\right)\)
Vì theo đầu bài ta có: \(x+y+z=0\)nên ta có ( DPCM ) ..... học cho tốt nhé!
L
1
31 tháng 8 2018
a) \(\left(x-1\right)\left(x^2+x+1\right)=x\left(x^2+x+1\right)-\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1=x^3-1\) đpcm
b) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)=\left(x-y\right)\left[x\left(x^2+y^2\right)+y\left(x^2+y^2\right)\right]\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)\) đpcm
13 tháng 8 2020
a) Ta có: \(VP=x^2+y^2+z^2-2xy+2yz-2zx\)
\(=\left(x^2-xy-xz\right)+\left(y^2-xy+yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x-y-z\right)+y\left(y-x+z\right)+z\left(z-y-x\right)\)
\(=x\left(x-y-z\right)-y\left(x-y-z\right)-z\left(x-y-z\right)\)
\(=\left(x-y-z\right)\left(x-y-z\right)\)
\(=\left(x-y-z\right)^2=VT\)(đpcm)
b) Ta có: \(VP=x^2+y^2+z^2+2xy-2yz-2zx\)
\(=\left(x^2+xy-zx\right)+\left(y^2+xy-2yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x+y-z\right)+y\left(x+y-z\right)+z\left(z-y-x\right)\)
\(=\left(x+y-z\right)\left(x+y\right)-z\left(x+y-z\right)\)
\(=\left(x+y-z\right)\left(x+y-z\right)\)
\(=\left(x+y-z\right)^2=VT\)(đpcm)
c) Ta có: \(VP=x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)=VT\)(đpcm)
d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5=VP\)(đpcm)
x4 -y4 =(x3+x2y+xy2+y3)(x-y)
VP=(x3+x2y+xy2+y3)(x-y)
=x.(x3+x2y+xy2+y3)-y.(x3+x2y+xy2+y3)
=x4+x3y+x2y2+xy3-x3y-x2y2-xy3-y4
=x4-y4=VT
=> x4 -y4 =(x3+x2y+xy2+y3)(x-y)