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Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)^2+\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+2+x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2\left(x-2\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)\left(x^2+6x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+6x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+3\right)^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\sqrt{3}-3\\x=-\sqrt{3}-3\end{matrix}\right.\)
Ta có: \(\left(x-2\right)\left(x^2+6x+6\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3-\sqrt{3}\right)\left(x+3+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3+\sqrt{3}\\x=-3-\sqrt{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(3x+2-x+6\right)\left(3x+2+x-6\right)=0\\ \Leftrightarrow\left(2x+8\right)\left(4x-4\right)=0\\ \Leftrightarrow4\left(x+4\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)
\(\Leftrightarrow2x^2-2x-2x^2=-6\)
\(\Leftrightarrow x=3\)
b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow24x+10=0\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
\(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=2\end{matrix}\right.\)
Vậy...