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(3-12x)(x-1)+(12x-8)(x+2)+x2=52
3(x-1)-12x(x-1)+12x(x+2)-8(x+2)+x2=52
3x-3-12x2+12+12x2+24x-8x-16+x2=52
(3x+24x-8x)+(12-3-16)+(12x2-12x2+x2)=52
19x-7+x2=52
x(19-x)=52+7=59
mà 59 là số ng tố nên x rỗng
Vậy x E \(\theta\)
c) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)
1,\(x^4-x=0\\ ->x\left(x-1\right)\left(x^2+x+1\right)=0\\ ->\left(......\right)\)
2\(x^4-x^2=0\\ ->x^2\left(x^2-1\right)\\ ->x^2\left(x-1\right)\left(x+1\right)\\ ->......\)
3,\(x^5+x^2\\ ->x^2\left(x^3+1\right)\\ ->x^2\left(x+1\right)\left(x^2-x+1\right)\\ ->.......\)
4\(3x\left(x-20\right)-x+20=0->\left(3x-1\right)\left(x-20\right)=0->.....\)
d) \(2x^3+3x^2+3x+1=2x^3+x^2+2x^2+x+2x+1\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2-x+1\right)\)
https://coccoc.com/search/math#query=3(1-4x).(x-1)%2B4.(3x-2).(x%2B2)%2Bx2+%3D52++T%C3%ACm+x+