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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Bài làm
a) x3 - 12 + 3x2 - 4x
= ( x3 + 3x2 ) + ( -12 - 4x )
= x2( x + 3 ) - 4( x + 3 )
= ( x2 - 4 )( x + 3 )
b) 2x2 + 13x - 7
= 2x2 + 14x - 1x - 7
= ( 2x2 + 14x ) - ( x + 7 )
= 2x( x + 7 ) - ( x + 7 )
= ( 2x - 1 )( x + 7 )
# Học tốt #
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
a) co sai de ko
b)x3-2x2+4x2-8x+3x-6=x2(x-2)+4x(x-2)+3(x-2)=(x-2)(x2+4x+3)=(x-2)(x+3)(x+1)
c)x3-2x2+2x2-4x-3x+6=x2(x-2)+2x(x-2)-3(x-2)=(x-2)(x2+2x-3)=(x-2)(x+3)(x-1)
d)x3-3x2+x2-3x-2x+6=x2(x-3)+x(x-3)-2(x-3)=(x-3)(x2+x-2)=(x-3)(x+2)(x-1)
a) x2 - 3x + 2 = x2 - x - 2x + 2 = x( x - 1 ) - 2( x - 1 ) = ( x - 1 )( x - 2 )
b) 2x2 - x - 6 = 2x2 - 4x + 3x - 6 = 2x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 2x + 3 )
c) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
d) x2 + 8x + 7 = x2 + x + 7x + 7 = x( x + 1 ) + 7( x + 1 ) = ( x + 1 )( x + 7 )
e) 3x2 + 2x - 5 = 3x2 - 3x + 5x - 5 = 3x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 3x + 5 )
f) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )
a \(x^2-3x+2=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)
b, \(2x^2-x-6=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\)
c, \(x^2-5x-6=x^2+x-6x-6=\left(x+1\right)\left(x-6\right)\)
d, \(x^2+8x+7=x^2+x+7x+7=\left(x+1\right)\left(x+7\right)\)
e, \(3x^2+2x-5=3x^2-3x+5x-5=\left(x-1\right)\left(3x+5\right)\)
f, \(4x^2-3x-1=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
\(\text{a) }x^2+4x+3\\ =x^2+x+3x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+3\right)\left(x+1\right)\\ \)
\(\text{b) }x^2-13x+12\\ =x^2-x-12x+12\\ =\left(x^2-x\right)-\left(12x-12\right)\\ =x\left(x-1\right)-12\left(x-1\right)\\ =\left(x-12\right)\left(x-1\right)\\ \)
\(\text{c) }x^2+5x-6\\ =x^2-x+6x-6\\ =\left(x^2-x\right)+\left(6x-6\right)\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\\ \)
\(\text{d) }2x^2+3x-5\\ =2x^2-2x+5x-5\\ =\left(2x^2-2x\right)+\left(5x-5\right)\\ =2x\left(x-1\right)+5\left(x-1\right)\\ =\left(2x+5\right)\left(x-1\right)\\ \)
\(\text{e) }a^{m+3}-a^m+a-1\\ =\left(a^{m+3}-a^m\right)+\left(a-1\right)\\ =a^m\left(a^3-1\right)+\left(a-1\right)\\ =a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)\\ =\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]\\ =\left(a-1\right)\left(a^{m+2}+a^{m+1}+1\right)\\ \)
\(x^2+4x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+3\right)\left(x+1\right)\)
\(x^2-13x+12=x\left(x-12\right)-\left(x-12\right)=\left(x-12\right)\left(x-1\right)\)
\(x^2+3x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x-2\right)\left(x+5\right)\)
\(x^2+5x-6=x\left(x-6\right)+\left(x-6\right)=\left(x-6\right)\left(x+1\right)\)
\(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
\(a^{m+3}-a^m+a-1=a^m\left(a^3-1\right)+\left(a-1\right)=a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)=\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]=\left(a-1\right)\left(a^{m+2}+a^{m+1}+a^m+1\right)\)