a/ \(2x^3+5x^2-x-6=0\)
\(\Leftrightarrow2x^3-2x^2+7x^2-7x+6x-6=0\)
\(\Leftrightarrow2x^2\left(x-1\right)+7x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2+7x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2x^2+4x\right)+\left(3x+6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[2x\left(x+2\right)+3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy pt có 3 nghiệm............
b/ \(x^4-2x^3+4x^2-3x=10\)
\(\Leftrightarrow x^4-2x^3+4x^2-3x-10=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^2-8x\right)+\left(5x-10\right)=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x^2-x+5=0\end{matrix}\right.\)
+) Ta thấy: \(x^2-x+5=\left(x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\)
=> Vô nghiệm
=> \(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy pt có 2 nghiệm x = 2 hoặc x = -1
