Phân tích đa thức thành nhân tử :
(x+y+z)^2+(x+y-z)^2-4z^2
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AH
Akai Haruma
Giáo viên
12 tháng 7 2021
Lời giải:
$x^4y^4-z^4=(x^2y^2)^2-(z^2)^2=(x^2y^2-z^2)(x^2y^2+z^2)$
$=(xy-z)(xy+z)(x^2y^2+z^2)$
$(x+y+z)^2-4z^2=(x+y+z)^2-(2z)^2=(x+y+z-2z)(x+y+z+2z)$
$=(x+y-z)(x+y+3z)$
$\frac{-1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2=\frac{-4x^2+12xy-9y^2}{36}$
$=-\frac{4x^2-12xy+9y^2}{36}=-\frac{(2x-3y)^2}{36}=-\left(\frac{2x-3y}{6}\right)^2$
E
1
11 tháng 10 2020
Ta có: \(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=x\left(y-z\right)\left(y+z\right)+yz^2-x^2y+zx^2-y^2z\)
\(=x\left(y-z\right)\left(y+z\right)-\left(y^2z-yz^2\right)-\left(x^2y-zx^2\right)\)
\(=x\left(y-z\right)\left(y+z\right)-yz\left(y-z\right)-x^2\left(y-z\right)\)
\(=\left(y-z\right)\left(xy+zx-yz-x^2\right)\)
\(=\left(y-z\right)\left[\left(zx-yz\right)-\left(x^2-xy\right)\right]\)
\(=\left(y-z\right)\left[z\left(x-y\right)-x\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
DM
0
KL
1
S
27 tháng 7 2018
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left[\left(z-y\right)+\left(y-x\right)\right]\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-y\right)-x^2z^2\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2y^2-x^2z^2\right)+\left(z-y\right)\left(y^2z^2-x^2z^2\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left(-x^2y-x^2z+z^2y+z^2x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left[xz\left(z-x\right)+y\left(z-x\right)\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(z-y\right)\left(z-x\right)\left(xy+yz+xz\right)\)
NH
1
VX
10 tháng 11 2021
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)
Z
1
26 tháng 7 2017
\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=x\left(y^2-z^2\right)-y\left(y^2-z^2+x^2-y^2\right)+z\left(x^2-y^2\right)\)
\(=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
chúc bn hc tốt ^^
DN
1
4 tháng 10 2016
\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left[-\left(z^2-x^2\right)-\left(x^2-y^2\right)\right]+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=-x\left(z^2-x^2\right)+y\left(z^2-x^2\right)-x\left(x^2-y^2\right)+z\left(x^2-y^2\right)\)
\(=\left(z^2-x^2\right)\left(y-x\right)+\left(x^2-y^2\right)\left(z-x\right)\)
\(=\left(y-x\right)\left(z-x\right)\left(z+x\right)+\left(z-x\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(z-x\right)\left(x+y-z-x\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
DT
0
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
( x + y + z )2 + ( x + y - z )2 - 4z2
= [ ( x + y ) + z ]2 + [ ( x + y ) - z ]2 - 4z2 (1)
Đặt \(\hept{\begin{cases}x+y=a\\z=b\end{cases}}\)
(1) <=> ( a + b )2 + ( a - b )2 - 4b2
= a2 + 2ab + b2 + a2 - 2ab + b2 - 4b2
= 2a2 - 2b2
= 2( a2 - b2 )
= 2( a - b )( a + b )
= 2( x + y - z )( x + y + z )