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GT
4
J
22 tháng 10 2017
BẠn ơi , bạn đã có đáp án câu d chưa ? Mk cx đang thắc mắc câu đó nè. Nếu có đáp án thì cho mk xin nha
GT
2
TM
13 tháng 7 2017
a)\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^4-5x^2+4=x^4-4x^2-x^2+4=x^2\left(x^2-4\right)-\left(x^2-4\right)=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
c)\(\left(x+y+z\right)^3-x^3-y^3-z^3=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-\left(x^3+y^3\right)\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right)z+3z^2-\left(x^2-xy+y^2\right)\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)
\(=\left(x+y\right)\left(3xy+3xz+3yz+3z^2\right)\)
\(=\left(x+y\right)\left[3x\left(y+z\right)+3z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
d) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left(3x^2y+3xy^2+3xyz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
3 tháng 9 2018
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
20 tháng 2 2017
\(=\left(x^3+y^3\right)+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy+yz+zx\right)\)
TT
6
M
1 tháng 10 2017
\(x^3+y^3+z^3-3xyz=\left(x^3+y^3\right)-3xyz+z^3\)
\(=\left(x+y\right)^3-3xy.\left(x+y\right)-3xyz+z^3\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy.\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right).\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy.\left(x+y+z\right)\)
\(=\left(x+y+z\right).\left(x^2+y^2+z^2-zx-zy+2zy-3xy\right)\)
\(=\left(x+y+z\right).\left(x^2+z^2+y^2-zx-zy-xy\right)\)
Vừa làm xong . Chúc bạn học tốt !
1 tháng 10 2017
\(=\left(x+y\right)^3+z^z-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
HQ
1
27 tháng 10 2020
1) x3 + y3 + z3 - 3xyz
= ( x + y )3 - 3xy( x + y ) + z3 - 3xyz
= [ ( x + y )3 + z3 ) - [ 3xy( x + y ) + 3xyz ]
= ( x + y + z )[ ( x + y )2 - ( x + y )z + z2 ] - 3xy( x + y + z )
= ( x + y + z )( x2 + y2 + z2 + 2xy - xz - yz - 3xy )
= ( x + y + z )( x2 + y2 + z2 - xy - yz - xz )
2) Tạm thời đang bí chưa làm được :(
3) ( x2 - 2x )2( x2 - 2x - 1 ) - 6 ( đề có vấn đề -- )
4) x4 - 7x3 + 14x2 - 7x + 1
= x4 - 3x2 - 4x2 + x2 + 12x2 + x2 - 4x - 3x + 1
= ( x4 - 3x2 + x2 ) - ( 4x3 - 12x2 + 4x ) + ( x2 - 3x + 1 )
= x2( x2 - 3x + 1 ) - 4x( x2 - 3x + 1 ) + ( x2 - 3x + 1 )
= ( x2 - 3x + 1 )( x2 - 4x + 1 )
CC
2
1 tháng 11 2017
a,81-(x^2-4xy+4y^2)=81-(x-2y)^2=(9-(x-2y))(9+(x-2y))=(9-x+2y)(9+x-2y)
b,x^3+y^3+z^3-3xyz=(x^3+3(x^2)y+3x(y^2)+y^3)+z^3-3xyz-3xy(x+y)
=((x+y)^3+3((x+y)^2)z+3(x+y)z^2+z^3)-(3xyz-3xy(x+y))-3(x+y)z(x+y+z)
=(x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=(x+y+z)((x+y+z)^2-3(x+y)z-3xy)
=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3yz-3xz)=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
3 tháng 9 2018
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
NT
0
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)