Question

[Mai Nguyễn Bảo Ngọc]

phân tích đa thức thành nhân tử

a(b-c)^2+b(c-a)^2+c(a-b)^2-a^3-b^3-c^3+4abc

giải PT

(2x^2+x)^2-4(2x^2+x)+3=0

Answer 1

[qwerty]

Đặt \(2x^2+x=t\)

Ta có: \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)

\(\Rightarrow t^2-4t+3=0\)

\(\Leftrightarrow t=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot3}}{2\cdot1}\)

\(\Leftrightarrow t=\dfrac{4\pm\sqrt{16-12}}{2}\)

\(\Leftrightarrow t=\dfrac{4\pm\sqrt{4}}{2}\)

\(\Leftrightarrow t=\dfrac{4\pm2}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{4-2}{2}\\t=\dfrac{4+2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+x=3\\2x^2+x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{3}{2};x_2=-1;x_3=\dfrac{1}{2};x_4=1\)

Answer 2

[Lightning Farron]

bài 2:

Đặt \(t=2x^2+x\) thì ta có:

\(t^2-4t+3=0\)\(\Rightarrow\left(t-3\right)\left(t-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t-1=0\\t-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x^2+x=1\\2x^2+x=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x^2+x-1=0\\2x^2+x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)\left(2x-1\right)=0\\\left(x-1\right)\left(2x+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)