Question

[lê phúc]

Bài 2: Tìm x, biết:

a) 4x(x + 1) = 8( x + 1)                c) x2 – 6x + 8 = 0

b) x3 + x2 + x + 1 = 0                  d) x3 – 7x – 6 = 0

Answer 1

[Nguyễn Hoàng Minh]

\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

Answer 2

[Lấp La Lấp Lánh]

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

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