Question

[Duong Thi Minh]

giải chi tiết hộ mk:

 a)\(x^4-8x^2+x+12=0\)

b) \(x^4+5x^3-10x^2+10x+4=0\)

Answer 1

[Thắng Nguyễn]

a)\(x^4-8x^2+x+12=0\)

\(\Leftrightarrow x^4-x^3-3x^2+x^3-x^2-3x-4x^2+4x+12=0\)

\(\Leftrightarrow x^2\left(x^2-x-3\right)+x\left(x^2-x-3\right)-4\left(x^2-x-3\right)=0\)

\(\Leftrightarrow\left(x^2-x-3\right)\left(x^2+x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-x-3=0\\x^2+x-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\Delta\left(1\right)=\left(-1\right)^2-\left(-4\left(1\cdot3\right)\right)=13\\\Delta\left(2\right)=1^2-\left(-4\left(1\cdot4\right)\right)=17\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x_{1,2}=\frac{1\pm\sqrt{13}}{2}\\x_{1,2}=\frac{-1\pm\sqrt{17}}{2}\end{cases}}\)

b)\(x^4+5x^3-10x^2+10x+4=0\)

\(\Leftrightarrow x^4-2x^3+2x^2+7x^3-14x^2+14x+2x^2-4x+4=0\)

\(\Leftrightarrow x^2\left(x^2-2x+2\right)+7x\left(x^2-2x+2\right)+2\left(x^2-2x+2\right)=0\)

\(\Leftrightarrow\left(x^2-2x+2\right)\left(x^2+7x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x+2=0\\x^2+7x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\Delta\left(1\right)=\left(-2\right)^2-4\cdot1\cdot2=-4< 0\left(loai\right)\\\Delta\left(2\right)=7^2-4\cdot1\cdot2=41\end{cases}}\)\(\Rightarrow x_{1,2}=\frac{-7\pm\sqrt{41}}{2}\)

Answer 2

[Duong Thi Minh]

Cảm ơn b Thắng Nguyễn