phân tích đa thức sao thành nhân tử
x^3-6x^2+12x-8
phân tích đa thức thành nhân tử
x^2-6x+7
giúp em với ạ
\(=x^2-6x+9-2=\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)
\(=\left(x^2-6x+9\right)-2=\left(x-3\right)^2-\sqrt{2^2}=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)
Phân tích đa thức thành nhân tử
x^3-4x^2+8x-8
\(x^3-4x^2+8x-8=x^2\left(x-2\right)-2x\left(x-2\right)+4\left(x-2\right)=\left(x-2\right)\left(x^2-2x+4\right)\)
\(x^3-4x^2+8x-8\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+4\right)\)
phân tích đa thức thành nhân tử x^3+y^3+6x^2+12x+8
\(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
phân tích đa thức thành nhân tử x^3+y^3+6x^2+12x+8
=(x^3+6x^2+12x+8)+y^3
=(x^3+3x^2+3x2^2+2^3)+y^3
=(x+2)^3+y^3
=(x+2+y)((x+2)^2-(x+2)y+y^2)
=(x+2+y)(x^2+4x+4-xy-2y+y^2)
=(x+2+y)(x^2+y^2-xy+4x-2y+4)
ko bít đâu nha : lớp 5
phân tích đa thức thành nhân tử x^3+y^3+6x^2+12x+8
x3 - 6x2 + 12x - 8
= x3 - 2x2 - 4x2 + 4x + 8x - 8
= (x3 - 2x2) - (4x2 - 8x) + (4x - 8)
= x2.(x - 2) + 4x.(x - 2) + 4.(x - 2)
= (x - 2).(x2 + 4x + 4)
= (x - 2).(x2 + 2x + 2x + 4)
= (x - 2).[x.(x + 2) + 2.(x + 2)]
= (x - 2).(x + 2).(x + 2)
= (x - 2).(x + 2)2
những bạn nhầm đề bài r bạn
x^3+y^3+6x^2+12x+8
=(x^3+6x^2+12x+8)+y^3
=(x+2)^3+y^3
=(x+2+y).(x^2+4x+4+y^2-xy-2y)
\(x^3+y^3+6x^2+12x+8\)
em hãy phân tích đa thức trên thành nhân tử
\(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
\(x^3+y^3+6x^2+12x+8\)
=\(x^3+3.2.x^2+3.2^2.x+2^3+y^3\)
\(=\left(x+2\right)^3+y^3=\left(x+2+y\right)\left(\left(x+2\right)^2-\left(x+2\right)y+y^2\right)\)
\(=\left(x+y+2\right)\left(x^2+4x+4-xy-2y-y^2\right)\)
Giúp mình với ạ
Phân tích đa thức thành nhân tử
X^3-6x^2-24x+144
\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)
\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)
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
x³ - 9x² + 6x + 16
x³ - x² - x - 2
x³ + x² - x + 2
x³ - 6x² - x + 30
x² - 7x - 6
27x³ - 27x² + 18x - 4
2x³ - x² + 5x + 3
(x² - 3)² + 16
a: \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b: \(x^3-x^2-x-2\)
\(=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\cdot\left(x^2+x+1\right)\)
c: \(x^3+x^2-x+2\)
\(=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
d: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
e: Sửa đề: \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
f: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
g: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
h: \(\left(x^2-3\right)^2+16\)
\(=x^4-6x^2+9+16\)
\(=x^4-6x^2+25\)
\(=x^4+10x^2+25-16x^2\)
\(=\left(x^2+5\right)^2-\left(4x\right)^2\)
\(=\left(x^2+5+4x\right)\left(x^2+5-4x\right)\)