Phân tích đa thức thành nhân tử:
\(x^2+12x+36=0\)
\(4x^2-4x+1=0\)
\(x^3+6x^2+12x+8=0\)
K
Khách
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Những câu hỏi liên quan
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:
\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)
LT
5
CW
12 tháng 12 2016
\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1-x-3\right)=0\)
\(\Leftrightarrow-2\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\)
V2
1
25 tháng 12 2021
\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
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: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
HH
2
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) \(4x^2+12x+9=\left(2x+3\right)^2\)
LL
1
KT
30 tháng 7 2018
a) \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
c) \(\left(2y-1\right)^1-4x^2+4x-1=\left(2y-1\right)^2-\left(2x-1\right)^2=\left(2y-1-2x+1\right)\left(2y-1+2x-1\right)\)
\(=\left(2y-2x\right)\left(2y+2x-2\right)=4\left(y-x\right)\left(y+x-1\right)\)
21 tháng 7 2019
a)\(2x^2-12x=-18\)
\(\Leftrightarrow2x^2-12x+18=0\)
\(\Leftrightarrow2\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
b) \(\left(4x^2-4x+1\right)-x^2=0\)
\(\Leftrightarrow\left(2x-1\right)^2-x^2=0\)
\(\Leftrightarrow\left(2x-1-x\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
_Minh ngụy_
21 tháng 7 2019
\(x^2-ay-y^2-ax\)
\(=\left(x^2-y^2\right)-\left(ax+ay\right)\)
\(=\left(x-y\right)\left(x+y\right)-a\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-a\right)\)
_Minh ngụy_
KC
1
8 tháng 10 2021
a) \(8x\left(x-3\right)+x-3=0\)
\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)
b) \(x^2+36=12x\)
\(\Rightarrow x^2-12x+36=0\)
\(\Rightarrow\left(x-6\right)^2=0\)
\(\Rightarrow x=6\)
PT
2
23 tháng 12 2018
a)\(4x^2-4x+4=0\Leftrightarrow\left(2x-1\right)^2+3\) (đến đây hết pt dc rùi)
b)\(x^3-27=\left(x-3\right)\left(x^2+3x+9\right)\)
c)\(x^3-4x^2+3x=x^3-x^2-3x^2+3x\)
=\(x^2\left(x-1\right)-3x\left(x-1\right)\)
=\(x\left(x-3\right)\left(x-1\right)\)
d)\(4x^2-12x+3=\left(2x-3\right)^2-6\)
=\(\left(2x-3\right)^2-\sqrt{6^2}\)
=\(\left(2x-3-\sqrt{6}\right)\left(2x-3+\sqrt{6}\right)\)
HN
23 tháng 12 2018
\(a,4x^2-4x+4=4\left(x^2-x+1\right)\)
\(b,x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)
\(c,x^3-4x^2+3x=x\left(x^2-4x+3\right)\)
\(=x\left[\left(x^2-x\right)-\left(3x-3\right)\right]\)
\(=x\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(x-3\right)\)
\(d,4x^2-12x+3=4\left(x^2-3x+\frac{3}{4}\right)\)
\(=4\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}+\frac{3}{4}\right)\)
\(=4\left[\left(x-\frac{3}{2}\right)^2-\frac{3}{2}\right]\)
\(=4\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2\right]\)
\(=4\left(x-\frac{3}{2}-\frac{\sqrt{3}}{\sqrt{2}}\right)\left(x-\frac{3}{2}+\frac{\sqrt{3}}{\sqrt{2}}\right)\)
\(=4\left(x-\frac{3+\sqrt{6}}{2}\right)\left(x-\frac{3-\sqrt{6}}{2}\right)\)
P/s: Dương: câu d t k chắc nx, sai thì thông cảm :)) -Huyền Nhi-
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