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\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 )
\(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 )
MTC : ( x - 2 )( x - 3 )( x + 3 )
=> \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\)
b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\))
\(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 )
MTC : \(2\left(x-1\right)\left(2x-1\right)\)
=> \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\)
c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b )
MTC : \(\left(a-b\right)\left(a+b\right)\)
=> \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)
d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 )
\(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 )
MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)
=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)
Sai chỗ nào bạn bỏ qua nhé
a) \(3x^2-9x+30=3\left(x^2-3x+10\right)\)
b) \(3x^2-5x-2=3x^2-6x+x-2\)
\(=3x\left(x-2\right)+\left(x-2\right)=\left(3x+1\right)\left(x-2\right)\)
c) \(x^4+4y^4\)
\(=x^4+4y^4+2x^2y^2+2x^2y^2-4x^2y^2+4xy^3-4xy^3+2x^3y-2x^3y\)
\(=\left(4y^4-4xy^3+2x^2y^2\right)+\left(4xy^3-4x^2y^2+2x^3y\right)\)
\(+\left(2x^2y^2-2x^3y+x^4\right)\)
\(=2y^2\left(2y^2-2xy+x^2\right)+2xy\left(2y^2-2xy+x^2\right)\)
\(+x^2\left(2y^2-2xy+x^2\right)\)
\(=\left(2y^2+2xy+x^2\right)\left(2y^2-2xy+x^2\right)\)
d) \(x^5+x+1\)
\(=x^5+x+1+x^4-x^4+x^3-x^3+x^2-x^2\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
******************************************************
a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)
c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)
d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)
\(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\)
e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\)
\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
Bài làm
a) 3x2 - 6x2 + 3x
= -3x2 + 3x
= 3x( 1 - x )
b) 3x2 + 5x - 3xy - 5y
= ( 3x2 - 3xy ) + ( 5x - 5y )
= 3x( x - y ) + 5( x - y )
= ( x - y )( 3x + 5 )
c) x3 + 2x2 + x
= x( x2 + 2x + 1 )
= x( x2 + 2.x.1 + 12 )
= x( x + 1 )2
d) xy + y2 - x - y
= ( xy - x ) + ( y2 - y )
= x( y - 1 ) + y( y - 1 )
= ( y - 1 )( x + y )
# Học tốt #
A . 5(x-y)-y(x-y)
=(x6-y)(5-y)
B . x^2 - xy - 8x+8y
=(x^2-xy)-(8x-8y))
=x(x-y) - 8(x-y)
C. x^2-10x+25 - y^2
=(x^2 - 10x + 25 ) - y^2
=(x-5)^2 - y^2
=(x-5+y)(x-5-y)
D . x^3 - 3x^2-4x+12
=(x^3 - 3x^2 ) - (4x - 12)
=x^2 (x-3)-4(x-3)
=(x^2-4)(x-3)
=(x+2)(x-2)(x-3)
D . 2x^2-2y^2- 6x-6y
=(2^x - 2y^2) - (6x+ 6y)
=2(x^2 - y^2) - 6(x+y)
=2(x+y)(x-y) - 6(x+y)
=2(x+y)(x-y-3)
E . x^3 - 3x^2 + 3x - 1
=(x-1)^3
D.x^2+3x+2
=x^2+2x+x+2
=(x^2+2x)+(x+2)
=x(x+2)+(x+2)
=(x+2)(x+1)
a, \(x^3+3x^2+6x+4\)
\(=x^3+x^2+2x^2+2x+4x+4\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x^2+2x+4\right)\left(x+1\right)\)