a) \(x^4+2x^3-12x^2-13x+42=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)
\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
Ta có:
\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy...........
b)\(\left(x-1\right)\left(x+2\right)\left(x+4\right)\left(x+7\right)=16\)
\(\Leftrightarrow\left(x-1\right)\left(x+7\right)\left(x+2\right)\left(x+4\right)=16\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x+8\right)-16=0\)
Đặt \(x^2+6x+0,5=t\)
\(\Rightarrow\left(t-7,5\right)\left(t+7,5\right)-16=0\)
\(\Rightarrow t^2-56,25-16=0\)
\(\Leftrightarrow t^2-72,25=0\)
\(\Leftrightarrow\left(t-\dfrac{17}{2}\right)\left(t+\dfrac{17}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{17}{2}\\t=-\dfrac{17}{2}\end{matrix}\right.\)
Xét \(t=\dfrac{17}{2}\Rightarrow x^2+6x+0,5=\dfrac{17}{2}\)
\(\Leftrightarrow x^2+6x-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
TT xét \(t=-\dfrac{17}{2}\Rightarrow x=\)-3
