a) 2x3+5x2-3x=0
<=> 2x3+6x2-x2-3x=0
<=> 2x2(x+3)-x(x+3)=0
<=> (x+3)(2x2-x)=0
<=> (x+3)x(2x-1)=0
\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
c) x3+1=x(x+1)
<=> (x+1)(x2+1-x)-x(x+1)=0
<=> (x+1)(x2-2x+1)=0
<=> (x+1)(x-1)2=0
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy ...
Hình như bạn ghi sai đề, nếu pt bậc 3 thì Carnado hoặc các cách khác, mình sửa lại đề
\(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(x^2+11x^2+15x-14-4=0\)
\(12x^2+15x-18=0\)
\(12x^2+24x-9x-18=0\)
\(12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\left(12x-9\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}12x-9=0\Leftrightarrow x=\dfrac{3}{4}\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)
