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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)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a,\(8x^2-8xy+2x=2x\left(4x-8y+1\right)\)
b,\(\left(x^2+2x\right)\left(x^2+4x+3\right)-24=x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24=\left(t+1\right)\left(t-1\right)-24=t^2-5^2=\left(t+5\right)\left(t-5\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)( đặt t = x2 + 3x + 1 )
a) \(\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-y-2\right)\left(x+y\right)\)
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
1:
a) \(x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
2
\(-2x^2-4x+6=0\)
\(\Leftrightarrow-2\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-3\end{array}\right.\)
1,
a) x( x2 + 2x +1) = x(x+1)2
b)25 - (x-2y)2 = (5-x+2y)(5+x-2y)
2,
(x-1)(x+3)=0
<=>x=1 hoặc x=-3
x3+2x2+x
= x ( x\(^2\) + 2x + 1 )
= x ( x + 1 )\(^2\)