a) \(x^2+x=6\)
\(\Rightarrow x^2+x-6=0\)
\(\Rightarrow x^2-2x+3x-6=0\)
\(\Rightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x+3=0\)
1) x-2 =0 => x=2
2) x+3=0 => x=-3
vậy x=2 ; x=-3
b)
\(6x^3+x^2=2x\)
\(\Rightarrow6x^3+x^2-2x=0\)
\(\Rightarrow6x^3-3x^2+4x^2-2x=0\)
\(\Rightarrow\left(6x^3-3x^2\right)+\left(4x^2-2x\right)=0\)
\(\Rightarrow3x^2\left(2x-1\right)+2x\left(2x-1\right)=0\)
\(\Rightarrow x\left(2x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow x=0\) hoặc \(2x-1=0\) hoặc \(3x+2=0\)
1) x=0
2) 2x-1=0 => 2x=1=> x=\(\dfrac{1}{2}\)
3) 3x+2=0 => 3x=-2 => x= \(-\dfrac{2}{3}\)
Vậy \(x=0\); \(x=\dfrac{1}{2}\);\(x=-\dfrac{2}{3}\)
a) \(x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b) \(6x^3+x^2=2x\)
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left(\left(6x^2+4x\right)-\left(3x+2\right)\right)=0\)
\(\Leftrightarrow x\left(2x\left(3x+2\right)-\left(3x+2\right)\right)=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) từ b nội suy tự làm a
x =0 là nghiệm
x khác 0 chia hai vế cho x
<=> 6x^2 +x =2
\(\Leftrightarrow\left(\sqrt{6}x\right)^2+2\dfrac{1}{\sqrt{6}}\left(\sqrt{6}x\right)=2\)
\(\left(\sqrt{6}x\right)^2+2\dfrac{1}{2\sqrt{6}}\left(\sqrt{6}x\right)+\left(\dfrac{1}{2\sqrt{6}}\right)^2=2+\dfrac{1}{24}=\dfrac{49}{4.6}\)
\(\left(\left(\sqrt{6}x\right)+\dfrac{1}{2\sqrt{6}}\right)^2=\left(\dfrac{7}{2\sqrt{6}}\right)^2\)
\(\left|\left(\sqrt{6}x\right)+\dfrac{1}{2\sqrt{6}}\right|=\dfrac{7}{2\sqrt{6}}\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{7-1}{2\sqrt{6}.\sqrt{6}}=\dfrac{1}{2}\\x_2=\dfrac{-7-1}{2\sqrt{6}\sqrt{6}}=\dfrac{-2}{3}\end{matrix}\right.\)