Tìm x biết 9x+2+9x-92.82=0
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Ta có: \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
hay \(x\in\left\{0;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5-x\right)\left(9x^2-4\right)=0\)
=>\(\left(x-5\right)\left(3x-2\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\3x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left(5-x\right)\left(9x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)
\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)
Vậy ...
\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)
\(\text{Vì }9x^2\ge0\)
\(\Rightarrow9x^2+9\ge9\)
\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x.\left(x^2+1\right)+9.\left(x^2+1\right)=0\\x.\left(x^2+1\right)-9.\left(x^2+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>|x^3+x|=|9x^2+9|
=>x^3+x=9x^2+9 hoặc x^3+x=-9x^2-9
=>x^3-9x^2+x-9=0 hoặc x^3+9x^2+x+9=0
=>x+9=0 hoặc (x-9)(x^2+1)=0
=>x=9 hoặc x=-9
![](https://rs.olm.vn/images/avt/0.png?1311)
9x2-6x-3=0
=>9x2-9x+3x-3=0
=>(x-1)(9x-3)=0
=>x-1=0 hoặc 9x+3 = 0
=> x=1 hoặc x=-1/3
b. x3+9x2+27x+19=0
x3+x2+8x2+8x+19x+19=0
(x+1)(x2+8x+19)=0
x+1=0 => x=-1
x2+8x+19= x2+8x+16+3=(x+4)2+3 lớn hơn hoặc bằng 3., lớn hơn 0 với moị x
a, \(\Rightarrow3\left(3x^2-2x-1\right)=0\)
\(\Rightarrow3x^2-2x-1=0\)
\(\Rightarrow x\left(3x-2\right)=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\3x-2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\3x-2=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)
b,\(\Rightarrow x^3+3x^2+6x^2+9x+18x+19=0\)
\(\Rightarrow x^2\left(x+3\right)+3x\left(x+3\right)+18\left(x+3\right)-2=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+3x+18\right)=2\)
Mk k co thoi gian. buoc tiep theo tu lam not nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-4x^2-9x+36=0\)
\(x^2\left(x-4\right)-9\left(x-4\right)=0\)
\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)
\(\Rightarrow x-4=0\) hay \(x^2-9=0\)
\(\Rightarrow x=4\) hay \(x^2=9=3^2\)
\(\Rightarrow x=4\) hay \(x=\pm3\)
⇔x2(x-4) -9(x-4) = 0
⇔(x-4).(x-3).(x+3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(4x^3-9x=0\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
Vậy x = 0 hoặc \(x=\frac{3}{2}\)
b) \(x^3+8x=0\Leftrightarrow x\left(x^2+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-8\left(L\right)\end{cases}}\)
Vậy x = 0
c) \(-x^3+9x=0\Leftrightarrow x\left(-x^2+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x^2+9=0\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=9\\x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)