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Đặt \(x^2-3x-1=a\)thay vào biểu thức ta được :
\(a^2-12a+27\)
\(=a^2-3a-9a+27\)
\(=a\left(a-3\right)-9\left(a-3\right)\)
\(=\left(a-3\right)\left(a-9\right)\)(1)
Thay \(a=x^2-3x-1\)vào (1) ta được :
\(\left(x^2-3x-1-3\right)\left(x^2-3x-1-10\right)\)
\(=\left(x^2-3x-4\right)\left(x^2-3x-11\right)\)
Bạn Châu sai đáp án cuối
phải là (x2-3x-4)(x2-3x-10) nha
\(3x^4+6x^3-7x^2+8x-10\)
\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)
\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)
Ta có (6x+5)2(3x+2)(x+1)-35
= (36x2+60x+25)(3x2+5x+2)-35 (1)
Đặt a=3x2+5x+2
=> 12a+1= 12(3x2+5x+2)+1 =36x2+60x+25
Thay a=3x2+5x+2 vào (1) ta được
(12a+1).a-35=12a2+a-35
= 12a2-20a+21a-35
= 4a(3a-5)+7(3a-5)
= (3a-5)(4a+7) (2)
Thay 3x2+5x+2=a vào (2) ta được
(9x2+15x+6-5)(12x2+20x+8+7)
= (9x2+15x+1)(12x2+20x+15)
Ta có: \(\left(6x+5\right)^2\left(3x+2\right)\left(x+1\right)-35\)
\(=\left(36x^2+60x+25\right)\left(3x^2+5x+2\right)-35\)(1)
Đặt \(3x^2+5x+2=y\)
\(\left(1\right)=\left(12y+1\right)y-35\)
\(=12y^2+y-35\)
\(=\left(3y-5\right)\left(4y+7\right)\)
\(=\left(9x^2+15x+1\right)\left(12x^2+20x+15\right)\)
3x3 + 6x2 + 3x - 12xy2
= 3x(x2 + 2x + 1 - 4y2)
= 3x[(x + 1)2 - (2y)2]
= 3x(x + 1 + 2y)(x - 2y + 1)
\(3x^3+6x^2+3x-12xy^2\)
\(=3x\left(x^2+2x+1-4y^2\right)\)
\(=3x\left[\left(x+1\right)^2-\left(2y\right)^2\right]\)
\(=3x\left(x+1-2y\right)\left(x+1+2y\right)\)
a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
Mình ko thêm bớt hạng tử nhé.
\(8x^3-3x+6x^2-1\)
\(=\left(8x^3-1\right)+\left(6x^2-3x\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right)+3x\left(2x-1\right)\)
\(=\left(2x-1\right)\left[\left(4x^2+2x+1\right)+3x\right]\)
\(=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(=\left(2x-1\right)\left[4x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(2x-1\right)\left(x+1\right)\left(4x+1\right)\)
\(8x^3-3x+6x^2-1=\left(8x^3-12x^2+6x-1\right)+\left(18x^2-9x\right)\)
\(=\left(\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\right)+\left(18x^2-9x\right)\)
\(=\left(2x-1\right)^3+9x\left(2x-1\right)=\left(2x-1\right)\left(\left(2x-1\right)^2+9x\right)\)
\(=\left(2x-1\right)\left(4x^2-4x+1+9x\right)=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
3x2 + 6x + 12
= 3(x2 + 2x + 4)
= 3(x + 2)2
thanks