Question

tìm x biết:

a) 4x (x+1) = 8(x+1)

b) (1-x)^2 -1+x=0

Answer 1

a, \(4x\left(x+1\right)=8.\left(x+1\right)\)

\(4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\left(x+1\right).\left(4x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\4x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\2\end{matrix}\right.\)

Vậy.........

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

\(\left(1-x\right).\left(1-x\right)-\left(1-x\right)=0\)

\(\left(1-x\right).\left(1-x-1\right)=0\)

\(\left(1-x\right).\left(-x\right)=0\)

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

Vậy.........

Answer 2

a)

\(4x\left(x+1\right)=8\left(x+1\right)\\ \Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\\ \Leftrightarrow\left(x+1\right)4\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy...

b)

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

Answer 3

a) 4x(x + 1) = 8(x + 1)

<=> 4x(x + 1) - 8(x + 1) = 0

<=> 4(x - 1)(x + 1) = 0

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

b) (1 - x)² - 1 + x = 0

<=> (1 - x)² - (1 - x) = 0

<=> (1 - x)(1 - x - 1) = 0

<=> -x(1 - x) = 0

<=> \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)