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Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
1) \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2) \(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x+2x+3x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
3) \(x^4+2x^2-3\)
\(=\left(x^2+1\right)^2-4\)
\(=\left(x^2+1-2\right)\left(x^2+1+2\right)\)
\(=\left(x^2-1\right)\left(x^2+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4) \(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
1, \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2, \(x^3+6x^2+11x+6\)
\(=\left(x^3+3x^2\right)+\left(3x^2+9x\right)+\left(2x+6\right)\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
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 1:
a: \(x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
b: \(x^3-6x^2-9x+14\)
\(=x^3-7x^2+x^2-7x-2x+14\)
\(=\left(x-7\right)\left(x^2+x-2\right)\)
\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)
c: \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)