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28 tháng 10 2018
a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)
\(=x\left(x+y-2z\right)\left(x+y+2z\right)\)
b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)
c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)
d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
e)\(x\left(2-x\right)-x+2=x\left(2-x\right)+\left(2-x\right)=\left(2-x\right)\left(x+1\right)\)
f)\(2x^2+3x-2=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g)\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)\)\(=\left(x-3\right)^2\left(x+3\right)\)
B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)
AK
10 tháng 7 2018
a )
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
b )
\(-5x^2-16x-3\)
\(=-5x^2-15x-x-3\)
\(=-5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(-5x-1\right)\left(x+3\right)\)
c )
\(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)-5\right]\)
d )
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
AK
10 tháng 7 2018
P/s : Mình bổ sung :
a )
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d )
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
DV
0
BN
2
TN
24 tháng 11 2016
a) x2 + xy - 5x - 5y
=x(x+y)-5(x+y)
=(x-5)(x+y)
b) x2 - y2 - 4x + 4
=(x2-4x+4)-y2
=(x-2)2-y2
=(x-2-y)(x-2+y)
27 tháng 8 2017
a) x2 + xy - 5x - 5y
= x ( x + y ) - 5 ( x + y )
= ( x + y ) ( x - 5 )
b) x2 - y2 - 4x + 4
= ( x2 - 4x + 4 ) - y2
= ( x - 2 )2 - y2
= ( x - 2 + y ) ( x - 2 - y )
22 tháng 12 2020
Bài 1:
a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)
b) \(\left(2x-3\right).\left(x+2\right)=2x^2+4x-3x-6=2x^2-x-6\)
Bài 2:
a) \(x^3-2x^2+x=x.\left(x^2-2x+1\right)=x.\left(x-1\right)^2\)
b) \(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x.\left(x-y\right)+2.\left(x-y\right)=\left(x-y\right).\left(x+2\right)\)
c) Đề sai.
5 tháng 6 2016
a) x2 + 4x – y2 + 4;
=x2+4x+4-y2
=(x+2)2-y2
=(x+2-y)(x+2+y)
b) 3x2 + 6xy + 3y2 – 3z2;
=3.(x2+2xy+y2)-3z2
=3.(x+y)2-3z2
=3.[(x+y)2-z2]
=3.(x+y-x)(x+y+z)
c) x2 – 2xy + y2 – z2 + 2zt – t2.
=(x-y)2-(z2-2zt+t2)
=(x-y)2-(z-t)2
=[(x-y)-(z-t)][(x-y)+(z-t)]
=(x-y-z+t)(x-y+z-t)
a)
\(9x^2-3x+2y-4y^2\\=(9x^2-4y^2)-(3x-2y)\\=[(3x)^2-(2y)^2]-(3x-2y)\\=(3x-2y)(3x+2y)-(3x-2y)\\=(3x-2y)(3x+2y-1)\)
b)
\(3x^2-6xy+3y^2-5x+5y\\=3(x^2-2xy+y^2)-5(x-y)\\=3(x-y)^2-5(x-y)\\=(x-y)[3(x-y)-5]\\=(x-y)(3x-3y-5)\\Toru\)