phân tích đa thức thành nhân tử
\(x^2+4x+4\)
\(x^2+2x+1\)
\(4x^2+12x+9\)
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VN
2
TA
30 tháng 7 2021
a) `x^4+2x^3-4x-4`
`=(x^4-4)+(2x^3-4x)`
`=(x^2-2)(x^2+2)+2x(x^2-2)`
`=(x^2-2)(x^2+2+2x)`
b) `x^3-4x^2+12x-27`
`=(x^3-27)-(4x^2-12x)`
`=(x-3)(x^2+3x+9)-4x(x-3)`
`=(x-3)(x^2+3x+9-4x)`
`=(x-3)(x^2-x+9)`
c) `xy-4y-5x+20`
`=y(x-4)-5(x-4)`
`=(y-5)(x-4)`
30 tháng 7 2021
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+2x^3-4x\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c) Ta có: \(xy-4y-5x+20\)
\(=y\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(y-5\right)\)
HN
1 tháng 11 2019
1) \(x^4-2x^3+3x^2-2x+1\)
\(=x^2\left(x^2-x+1\right)-x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)^2\)
2) \(x^4-4x^3+10x^2-12x+9\)
\(=x^2\left(x^2-2x+3\right)-2x\left(x^2-2x+3\right)+3\left(x^2-2x+3\right)\)
\(=\left(x^2-2x+3\right)^2\)
NT
11 tháng 8 2021
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. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
12 tháng 8 2021
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: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
18 tháng 7 2021
a) \(4x^2+20x+25=\left(2x+5\right)^2\)
b) \(x^2-6x+9=\left(x-3\right)^2\)
c) \(4x^2+12x+9=\left(2x+3\right)^2\)
PD
2
LT
24 tháng 8 2016
x^2 - 2x - 15
= x^2 - 5x + 3x - 15
= ( x^2 + 3x ) - (5x +15 )
= x ( x +3 ) - 5 ( x + 3 )
(x + 3 ) ( x - 5 )
12 tháng 12 2023
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
5 tháng 7 2017
a) \(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
c) \(36-12x+x^2=x^2-12x+36=x^2-6x-6x+36\)
\(=x\left(x-6\right)-6\left(x-6\right)=\left(x-6\right)\left(x-6\right)=\left(x-6\right)^2\)
5 tháng 7 2017
\(x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(4x^2+12x+9\)
\(=\left(2x\right)^2+2.2x.3+9\)
\(=\left(2x+3\right)^2\)
\(36-12x+x^2\)
\(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
14 tháng 11 2021
Bài 1:
\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)
Bài 2:
\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)
4 tháng 2 2022
a: \(x^3-2x+4\)
\(=x^3+2x^2-2x^2-4x+2x+4\)
\(=\left(x+2\right)\left(x^2-2x+2\right)\)
b: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c: \(x^3+2x^2+2x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
25 tháng 12 2021
\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
x2 + 4x + 4 = (x + 2)2
x2 + 2x + 1 = (x + 1)2
4x2 + 12x + 9 = (2x + 3)2
\(x^2+4x+4=x\left(x+2\right)^2\)
\(x^2+2x+1=x\left(x+1\right)^2\)
\(4x^2+12x+9=\left(2x+3\right)^2\)