Tìm x, biết :
a, x^3+3x^2+2 = 0
b, x^3-12x^2+48x-72 = 0
Giúp mk vs ạ
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+2=0
hay x=-2
Vậy: x=-2
b) Ta có: \(x^3-12x^2+48x-72=0\)
\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)
\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)
mà \(x^2-6x+12>0\forall x\)
nên x-6=0
hay x=6
Vậy: x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1\ne0\forall x\)
nên x+2=0
hay x=-2
Vậy: x=-2
b) Ta có: \(x^3-12x^2+48x-72=0\)
\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)
\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)
mà \(x^2-6x+12\ne0\forall x\)
nên x-6=0
hay x=6
Vậy: x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
câu c,d,e lạ quá nhưng thui you viết thế nào tui làm thế ấy sai đừng trách
a) 5x(x-1)=x-1
<=> 5x(x-1)-(x-1)=0
<=>(x-1)(5x-1)=0
<=>x-1=0 hoặc 5x-1=0
<=>x=1 hoặc \(\frac{1}{5}\)
b) 2(x+5)-x*2-5x=0
VT=-5(x-2)
<=>-5(x-2)=0
<=>x-2=0
<=>x=2
c)(2x-3)*2-(x-5)*2=0
VT=2(x+2)
<=>2(x+2)=0
<=>x+2=0
<=>x=-2
d) 3x*3-48x=0
VT=-39x
<=>-39x=0
<=>x=0
e) x*3+x*2-4x=4
VT=x
<=>x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x-\frac{1}{2}\right|+1=3x\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) <=> x2 - 4x - x2 + 8 = 0 <=> x2 - 4x + 8 = 0
Dễ thấy phương trình vô nghiệm vì x2 - 4x + 8 = ( x - 2 )2 + 4 > 0
2) <=> ( x - 1 )3 = 0 <=> x = 1
3) <=> ( x - 2 )3 = 0 <=> x = 2
4) <=> ( 2x - 1 )3 = 0 <=> x = 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)
\(\Rightarrow x-3=-2\Leftrightarrow x=1\) vậy \(x=1\)
b) \(64x^3+48x^2+12x+1=0\Leftrightarrow\left(4x+1\right)^3=3^3\)
\(\Rightarrow4x+1=3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
c) \(x^3-3x^2+3x-1=0\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
d) \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\) vậy \(x=-2\)
e) \(x^3-6x^2+12x-8=0\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\) vậy \(x=2\)
a, Thắc mắc đề cóa sai khong
.
( đáp án vẫn có nhưng là số vô tỉ nên nghe lạ á )
b, Ta có : \(x^3-12x^2+48x-72=0\)
=> \(x^3-3.x^2.4+3.x.4^2-64-8=0\)
=> \(\left(x-4\right)^3-8=0\)
=> \(\sqrt[3]{\left(x-4\right)^3}=\sqrt[3]{8}=2\)
=> \(x=6\)
Vậy ....