Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

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Thùy Dung Hà

Tim x sao cho :

\(1+6x-6x^2-x^3=0\)

Quoc Tran Anh Le
1 tháng 7 2019 lúc 10:34

\(1+6x-6x^2-x^3=0\)

\(\Leftrightarrow-x^3-6x^2+6x+1=0\)

\(\Leftrightarrow-x^3+x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow-x^2\left(x-1\right)-7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x^2-7x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-x^2-7x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+12,25-11,25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5\right)^2-\left(\frac{3\sqrt{5}}{2}\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5-\frac{3\sqrt{5}}{2}\right)\left(x+3,5+\frac{3\sqrt{5}}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+3,5-\frac{3\sqrt{5}}{2}=0\\x+3,5+\frac{3\sqrt{5}}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3,5+\frac{3\sqrt{5}}{2}=\frac{-7+3\sqrt{5}}{2}\\x=-3,5-\frac{3\sqrt{5}}{2}=\frac{-7-3\sqrt{5}}{2}\end{matrix}\right.\)

Vậy x = \(\left\{1;\frac{-7+3\sqrt{5}}{2};\frac{-7-3\sqrt{5}}{2}\right\}\)

Trần Thanh Phương
1 tháng 7 2019 lúc 10:35

\(1+6x-6x^2-x^3=0\)

\(\Leftrightarrow x^3+6x^2-6x-1=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2-\frac{45}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2=\left(\frac{\pm\sqrt{45}}{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\pm\sqrt{45}-7}{2}\end{matrix}\right.\)