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a x2 - 7x - 6 = x2 -6x-x-6
= x(x-6)+(x-6)
= (x+1)(x+6)
b x3 + 4x2 - 7x -10
=x3 - 2x2 + 6x2 -12x+5x-10
= x2(x-2) + 6x(x-2) + 5(x-2)
= (x2 +6x+5)(x-2)
=(x2 +x+5x+5)(x-2)
=[x(x+1)+5(x+1)](x-2)
=(x+5)(x+1)(x-2)
c x3 +x2-2
=x3-x2+2x2-2x+2x-2
=x2(x-1)+2x(x-1)+2(x-1)
=(x2+2x+2)(x-1)
d x2 +5x+6
=x2 +2x+3x+6
=x(x+2)+3(x+2)
=(x+3)(x-2)
Con 1 y e nhung luoi lam
mk chỉnh lại đề
\(x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
x2-7x+12
=x2-3x-4x+12
=x(x-3)-4(x-3)
=(x-3)(x-4)
x4-4x2+4x-1
=x4-1-4x2+4x
=(x2-1)(x2+1)-4x(x-1)
=(x-1)(x+1)(x2+1)-4x(x-1)
=(x-1)[(x+1)(x2+1)-4x]
=(x-1)(x3+x2+x+1-4x)
=(x-1)(x3+x2-3x+1)
6x4-11x2+3
=6x4-2x2-9x2+3
=2x2(3x2-1)-3(3x2-1)
=(3x2-1)(2x2-3)
b,(x2-11x+28)(x2-7x+10)-72
(x-4)(x-7)(x-2)(x-5)-72
(x-2)(x-7)(x-4)(x-5)-72
(x2-9x+14)(x2-9x+20)-72
Đặt"t"=x2-9x+14, ta có:
(=)t(t+6)-72
(=)t2+6t-72
(=)t2+12t-6t-72
(=)t(t-6)-12(t-6)
(=)(t-6)(t-12)
(=)(x2-9x+14-6)(x2-9x+14-12)
(=)(x2-9x+8)(x2-9x+2)
c: \(=4x^3-3x^2+4x^2-3x+4x-3=\left(4x-3\right)\left(x^2+x+1\right)\)
b: \(=\left(x-4\right)\left(x-7\right)\left(x-2\right)\left(x-5\right)-72\)
\(=\left(x^2-9x+20\right)\left(x^2-9x+14\right)-72\)
\(=\left(x^2-9x\right)^2+34\left(x^2-9x\right)+208\)
\(=\left(x^2-9x+26\right)\left(x^2-9x+8\right)\)
\(=\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)
6x3 - 11x2 - x - 2
= 6x3 - 12x2 + x2 - 2x + x - 2
= ( 6x3 - 12x2 ) + ( x2 - 2x ) + ( x - 2 )
= 6x2( x - 2 ) + x( x - 2 ) + 1( x - 2 )
= ( x - 2 )( 6x2 + x + 1 )
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)