Phân tích đa thứ thành nhân tử f(x) =9x^2+12x-5
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a) \(27+x^3=\left(x+3\right)\left(x^2-3x+9\right)\)
b) \(-x^3+12x^2-48x+64=\left(4-x\right)^3\)
c) \(27+27x+9x^2=9\left(x^2+3x+3\right)\)
a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)
c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)
\(=\left(12x+9x^2+4\right)-\left(6y\right)^2=\left(3x+2\right)^2-\left(6y\right)^2\)
\(=\left(3x+2-6y\right)\left(3x+2+6y\right)\)
k mình cái
12x+9x2+4-36y2
= (9x2+12x+4)-36y2
= (3x+2)2-36y2
= ((3x+2)-6y2)((3x+2)+6y2)
=(3x+2-6y2)(3x+2+6y2)
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
\(=9x^3-3x^2-9x^2+6x-1\)1
\(=3x^2\left(3x-1\right)-\left(9x^2-6x+1\right)\)
\(=3x^2\left(3x-1\right)-\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(3x^2-3x+1\right)\)
a, f(x)= (x^5-x^4)-(4x^4-4x^3)+(5x^3-5x^2)-(4x^2-4x)+(4x-4)
=x^4(x-1)-4x^3(x-1)+5x^2(x-1)-4x(x-1)+4(x-1)
=(x^4-4x^3+5x^2-4x+4)(x-1)
=[(x^4-2x^3)-(2x^3-4x^2)+(x^2-2x)-(2x-4)](x-1)
=(x^3-2x^2+x-2)(x-2)(x-1)
=(x^2+1)(x-2)^2(x-1)
để mik kb cho.mik cũng đang buồn nè. nhớ đồng ý nhaaaaa