\(x^4+\left(x-1\right)\left(x^2-2x+2\right)=0\)
\(\Leftrightarrow x^4+x^3-3x^2+4x-2=0\)
\(\Leftrightarrow\left(x^4-x^3+x^2\right)+\left(2x^3-2x^2+2x\right)+\left(-2x^2+2x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+2x-2\right)=0\)
Ta có: \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)nên
\(\Rightarrow x^2+2x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)
Còn ít nhất 2 cách giải nữa. Các bạn vẫn còn cơ hội. Đây chỉ là một phương trình đưa ra về phương trình bậc hai thôi.
Cách khác:
\(x^4+\left(x-1\right)\left(x^2-2x+2\right)=0\)
\(\Leftrightarrow x^4+x^3-3x^2+4x-2=0\)
\(\Leftrightarrow4x^4+4x^3-12x^2+16x-8=0\)
\(\Leftrightarrow\left(4x^4+4x^3+x^2\right)-2\left(2x^2+x\right)+1=9x^2-18x+9\)
\(\Leftrightarrow\left(2x^2+x-1\right)^2=\left(3x-3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+x-1=3x-3\\2x^2+x-1=-3x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+2=0\\2x^2+4x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\left(l\right)\\x^2+2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)
\(x^4+(x-1)(x^2-2x+2)=0\)
\(\Leftrightarrow\)\(x ^4 + ( x − 1 ) ( x ^2 − 2 ( x − 1 ) ) =0\)
\(\Leftrightarrow\)\(x^4+(x-1)x^2 -2(x-1)^2=0\)
Đặt \(y=x-1\) , ta có:
\(x^4 + yx^2-xy^2=0\)
\(⇔2y^2−x^2y−x^4=0(1) \)
Xem (1) là phương trình bậc 2 theo y . Do đó dấu bằng xảy ra
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}y=x^2\\y=-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=x^2\\x-1=-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x^2+2x-2=0\end{matrix}\right.\) \(\Leftrightarrow x=-1\pm\sqrt{3}\)
Thôi 2 cách thôi. Nhường các bạn khác nữa :))
x4+(x−1)(x2−2(x−1))=0⇔x4+(x−1)x2−2(x−1)2=0" id="MathJax-Element-4-Frame" role="presentation" style="display: inline-block; line-height: 0; font-size: 15.82px; word-wrap: normal; word-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;" tabindex="0">x4+(x−1)(x2−2(x−1))=0⇔x4+(x−1)x2−2(x−1)2=y=x−1" id="MathJax-Element-5-Frame" role="presentation" style="display: inline-block; line-height: 0; font-size: 15.82px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative;" tabindex="0">: x4+yx2−2y2=0⇔2y2−x2y−x4=0
y" id="MathJax-Element-8-Frame" role="presentation" style="display: inline-block; line-height: 0; font-size: 15.82px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative;" tabindex="0">. Nên do đó ta được quyền suy ra: x4+yx2−2y2=0⇔2y2−x2y−x4=0 ⇔
(1)⇔[y=x2y=−x22⇔[x−1=x2x−1=&#x
