phân tích thành nhân tử
8x^3-12x^2+6x-1
Phân tích đa thức thành nhân tử
8x^3 + 6x^2 + 3x + 1
\(=\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\)
\(=\left(2x+1\right)\left(4x^2+x+1\right)\)
Phân tích đa thức thành nhân tử
8x^3 - 6x^2 + 3x - 1
\(=8x^3-4x^2-2x^2+x+2x-1=\left(2x-1\right)\left(4x^2-x+1\right)\)
Phân tích thành nhân tử
`2x-1^3 +8`
`8x^3 -12x^2 +6x-1`
`8x^3 -12x^2 +6x-2`
`9x^3 -12x^2 +6x-1`
\(2x-1^3+8\)
\(=2x-9\)
\(=\left(\sqrt{2x}\right)^2-3^2\)
\(=\left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)\)
_________
\(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
_______________
\(8x^3-12x^2+6x-2\)
\(=8x^3-12x^2+6x-1-1\)
\(=\left(2x-1\right)^3-1\)
\(=\left(2x-1-1\right)\left(4x^2-4x+1+2x-1+1\right)\)
\(=\left(2x-2\right)\left(4x^2-2x+1\right)\)
\(=2\left(x-1\right)\left(4x^2-2x+1\right)\)
________
\(9x^3-12x^2+6x-1\)
\(=x^3+8x^3-12x^2+6x-1\)
\(=x^3+\left(2x-1\right)^3\)
\(=\left(x+2x-1\right)\left(x^2-2x^2-x+4x^2-4x+1\right)\)
\(=\left(3x-1\right)\left(3x^2-5x+1\right)\)
b: 8x^3-12x^2+6x-1
=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3
=(2x-1)^3
c: =(8x^3-12x^2+6x-1)-1
=(2x-1)^3-1
=(2x-1-1)[(2x-1)^2+2x-1+1]
=2(x-1)(4x^2-4x+1+2x)
=2(x-1)(4x^2-2x+1)
8x³ - 12x² + 6x - 1
= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³
= (2x - 1)³
--------------------
8x³ - 12x² + 6x - 2
= 8x³ - 12x² + 6x - 1 - 1
= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³
= (2x - 1)³ - 1³
= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]
= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)
= 2(x - 1)(4x² - 2x + 1)
--------------------
9x³ - 12x² + 6x - 1
= x³ + 8x³ - 12x² + 6x - 1
= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³
= x³ + (2x - 1)³
= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]
= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)
= (3x - 1)(3x² - 3x + 1)
Phân tích đa thức thành nhân tử
\(-6xy^2+6x^3+12x^2+6x\)
\(=6x\left(-y^2+x^2+2x+1\right)\\ =6x\left[\left(x^2+2x+1\right)-y^2\right]\\ =6x\left[\left(x+1\right)^2-y^2\right]\\ =6x\left(x+1-y\right)\left(x+1+y\right)\)
Phân tích đa thức thành nhân tử:
\(x^2+12x+36=0\)
\(4x^2-4x+1=0\)
\(x^3+6x^2+12x+8=0\)
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
Phân tích đa thức thành nhân tử: 1, x^3+2x^2-6x-27 2, 9x^2+6x-4y^2-4y 3, 12x^3+4x^2-27x-9
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: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
Phân tích đa thức thành nhân tử
x\(^3\)-6x\(^2\)+12x-7
\(=\left(x^3-6x^2+12x-8\right)+1\\ =\left(x-2\right)^3+1\\ =\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)\\ =\left(x-1\right)\left(x^2-5x+7\right)\)
Phân tích đa thức thành nhân tử
27x^3+27x^2+9x+1
-x^3-3x^2-3x-1
- 8+12x-6x^2+x^3
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) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)
Phân tích thành nhân tử
(12x^2+6x)*(y+z)+(12x^2+6x)*(y-z)