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\(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2\)
= \(\left(x^2-1\right)^2-2.\left(x^2-1\right).\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{4}-2x^2\)
= \(\left(x^2-1-\frac{x}{2}\right)^2-\frac{9}{4}x^2\)
\(=\left(x^2-1-\frac{x}{2}-\frac{3}{2}x\right)\left(x^2-1-\frac{x}{2}+\frac{3}{2}x\right)\)
= \(\left(x^2-2x-1\right)\left(x^2-x-1\right)\)
Phân tích tiếp được đấy:
\(x^2-2x-1=\left(x-1\right)^2-2=\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
\(x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)
Thay vào nhé!
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)
\(\left(x+3\right)^2-\left(2x+6\right)\left(1-3x\right)+\left(3x+1\right)^2\)
\(=x^2+6x+9-\left(2x-6x^2+6-18x\right)+9x^2+6x+1\)
\(=10x^2+12x+10+6x^2+16x-6=16x^2+28x+4\)
\(=4\left(4x^2+7x+1\right)\)
a) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
\(=-\left(x+1\right)\left(2x^2-3x-7\right)\)
b) \(\left(x+y\right)\left(2x-y\right)-\left(3x-y\right)\left(y-2x\right)\)
\(=\left(x+y\right)\left(2x-y\right)+\left(3x-y\right)\left(2x-y\right)\)
\(=\left(2x-y\right)\left(x+y+3x-y\right)\)
\(=4x\left(2x-y\right)\)
c) \(5u\left(u-v\right)^2+10u^2\left(v-u\right)^2\)
\(=5u\left(u-v\right)^2+10u^2\left(u-v\right)^2\)
\(=5u\left(u-v\right)^2\left(1+2u\right)\)
Trả lời:
a, 3x ( x + 1 )2 - 5x2 ( x + 1 ) + 7 ( x + 1 )
= ( x + 1 )[ 3x ( x + 1 ) - 5x2 + 7 ]
= ( x + 1 )( 3x2 + 3x - 5x2 + 7 )
= ( x + 1 )( - 2x2 + 3x + 7 )
b, ( x + y )( 2x - y ) - ( 3x - y )( y - 2x )
= ( x + y )( 2x - y ) + ( 3x - y )( 2x - y )
= ( 2x - y )( x + y + 3x - y )
= 4x ( 2x - y )
c, 5u ( u - v )2 + 10u2 ( v - u )2
= 5u ( u - v )2 + 10u2 ( u - v )2
= 5u ( u - v )2( 1 + 2u )
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2\)
\(=x^4+1+2x^2+3x^2+3x+2x^2\)
\(=x^4+3x^3+4x^2+3x+1\)
\(=x^4+x^3+2x^3+2x^2+2x^2+2x+x+1\)
\(=\left(x+1\right)\left(x^3+2x^2+2x+1\right)\)
Đặt \(x^2+1=a\) thay vào ta được :
\(a^2+3ax+2x^2\)
\(=a^2+ax+2ax+2x^2\)
\(=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(a+2x\right)\left(a+x\right)\)
\(=\left(x^2+1+2x\right)\left(x^2+1+x\right)\)
\(=\left(x+1\right)^2\left(x^2+x+1\right)\)