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Phân tích đa thức thành nhân tử:
a) (x-1)(x-2)(x-3)(x-4)+1
b) (x2+3x+2)(x2+7x+12)+1
c) 12x2-3xy-8xz+2yz
a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)
\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)
\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)
Đặt \(a=x^2-5x+5\)
\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)
\(\Leftrightarrow A=a^2-1^2+1\)
\(\Leftrightarrow A=a^2\)
Thay \(a=x^2-5x+5\)vào A ta có :
\(A=\left(x^2-5x+5\right)^2\)
b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)
\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)
\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)
\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
Làm tương tự câu a)
c) \(12x^2-3xy-8xz+2yz\)
\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x-2z\right)\)
a) Ta có: \(27m\left(m+n\right)-m-n\)
\(=27m\left(m+n\right)-\left(m+n\right)\)
\(=\left(m+n\right)\left(27m-1\right)\)
b) Ta có: \(15x\left(x-y\right)-25x+25y\)
\(=15x\left(x-y\right)-25\left(x-y\right)\)
\(=5\left(x-y\right)\left(3x-5\right)\)
c) Ta có: \(12x^2-3xy+8xz-2yz\)
\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x+2z\right)\)
d) Ta có: \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
a) x3 + y3 - 3xy + 1
= ( x + y )3 - 3xy( x + y ) - 3xy + 1
= [ ( x + y )3 + 1 ] - [ 3xy( x + y ) + 3xy ]
= ( x + y + 1 )( x2 + 2xy + y2 - x - y + 1 ) - 3xy( x + y + 1 )
= ( x + y + 1 )( x2 - xy + y2 - x - y + 1 )
b) ( 4 - x )5 + ( x - 2 )5 - 32
= [ -( x - 4 ) ]5 + ( x - 2 )5 - 32
Đặt t = x - 3
đthức <=> ( 1 - t )5 + ( 1 + t )5 - 32 ( chỗ này bạn dùng nhị thức Newton để khai triển nhé )
= 10t4 + 20t2 - 30
Đặt y = t2
đthức = 10y2 + 20y - 30
= 10y2 - 10y + 30y - 30
= 10y( y - 1 ) + 30( y - 1 )
= 10( y - 1 )( y + 3 )
= 10( t2 - 1 )( t2 + 3 )
= 10( t - 1 )( t + 1 )( t2 + 3 )
= 10( x - 3 - 1 )( x - 3 + 1 )[ ( x - 3 )2 + 3 ]
= 10( x - 4 )( x - 2 )( x2 - 6x + 12 )
a,\(x^3+y^3-3xy+1\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy\)
\(=\left[\left(x+y\right)^3+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)\)
1. 12x\(^2\) - 3xy + 8xz - 2yz
= ( 12x\(^2\) - 3xy ) + ( 8xz - 2yz )
= 3x(4x - y) + 2z(4x - y)
= (4x - y)(3x+2z)
2. 4x\(^2\) - y\(^2\) +8(y-2)
= 4x\(^2\) - y\(^2\) + 8y -16
= 4x\(^2\) - ( y\(^2\) - 8y + 16)
= 4x\(^2\) - ( y - 4 )\(^2\)
= (2x-y+4)(2x+y-4)
a) ( x-3y ) ( x + 1 )
b) ( x+y+5 ) ( x+y-5 )
c) ( x-5 ) ( x+2 )
Hk tốt
a, \(12x^2-3xy-8xz+2yz=3x\left(4x-y\right)-2z\left(4x-y\right)=\left(4x-y\right)\left(3x-2z\right)\)
b: =(x^2+x)^2+3(x^2+x)+2-12
=(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)