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a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(x^3+3x^2+3x-7=0\)
\(\Leftrightarrow x+1=2\)
hay x=1
b: Ta có: \(x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)
Vậy .............................
b) \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)
\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)
Vậy ................................
c) \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)
\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)
\(\left(x^2-4\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)
a)\(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-4x=0\)
\(x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
b) \(4x^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\left(2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)
c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
d) \(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-2\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)
\(\left(x-3\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)
\(x^2-4x=0\)
\(x.\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)
\(4x^2-9=0\)
\(2^2x^2-9=0\)
\(\left(2x\right)^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\cdot\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
\(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-\left(4x+18\right)=0\)
\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)
\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(2\left(x+3\right)+x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x\left(3-\dfrac{1}{4}x\right)+\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow12x-x^2+x^2-4=0\Rightarrow12x=4\Rightarrow x=\dfrac{1}{3}\)
\(12x-x^2+x^2-2^2=0\)
\(12x-2=0\)
\(12x=2\)
\(x=\dfrac{1}{6}\)
Vậy x=1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
`x^3+4x^2+4x+3=0`
`<=>x^3+(x^2+3x^2)+(x+3x)+(1+3)=0`
`<=>(x^3+x^2+x)+(3x^2+3x+3)=0`
`<=>x(x^2+x+1)+3(x^2+x+1)=0`
`<=>(x+3)(x^2+x+1)=0`
`<=>x+3=0\ (x^2+x+1>0\ forall x)`
`<=>x=-3`