Question

Tìm x:

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

Answer 1

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\left(2x-\frac{1}{2}\right)\left(5x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\5x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{2}\\5x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{-2}{5}\end{cases}}\)

Answer 2

5x.(2x - 1/2) + 2.(2x - 1/2) = 0

<=> 5x.2x + 5x.(-1/2) + 2.2x + 2.(-1/2) = 0

<=> 10x² - 5/2x + 4x - 1 = 0

<=> 10x² - 13/2x - 1 = 0

=> x = 1/4 hoặc x = -2/5

Answer 3

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\left(5x+2\right)\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+2=0\\2x-\frac{1}{2}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}5x=0-2\\2x=0+\frac{1}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}5x=-2\\2x=\frac{1}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-2:5\\x=\frac{1}{2}:2\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{1}{4}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{1}{4}\end{cases}}\)

Answer 4

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

Answer 5

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

\(\left(5x+2\right).\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+2=0\\2x-\frac{1}{2}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-2}{5}\\x=\frac{1}{4}\end{cases}}\)