Question

Tìm x 

a) (2x-1)(2x+1)-4(x^2+x)=16

b) 5x(x-2013)-x+2013=0

Answer 1

a) ( 2x - 1 )( 2x + 1 ) - 4( x² + x ) = 16

⇔ 4x² - 1 - 4x² - 4x = 16

⇔ -4x - 1 = 16

⇔ -4x = 17

⇔ x = -17/4

b) 5x( x - 2013 ) - x + 2013 = 0

⇔ 5x( x - 2013 )  - ( x - 2013 ) = 0

⇔ ( x - 2013 )( 5x - 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

Answer 2

a) \(\left(2x-1\right)\left(2x+1\right)-4.\left(x^2+x\right)=16\)

\(4x^2-1-4x^2-4x=16\)

\(-1-4x=16\)

\(-4x=17\)

\(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\left(x-2013\right)\left(5x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2013=0\\5x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}}\)

Answer 3

a) \(\left(2x-1\right)\left(2x+1\right)-4\left(x^2+x\right)=16\)

\(\Leftrightarrow4x^2-1-4x^2-4x=16\)

\(\Leftrightarrow-1-4x=16\)\(\Leftrightarrow4x=-17\)\(\Leftrightarrow x=-\frac{17}{4}\)

Vậy \(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\Leftrightarrow5x\left(x-2013\right)-\left(x-2013\right)=0\)

\(\Leftrightarrow\left(x-2013\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

Vậy \(x=2013\)hoặc \(x=\frac{1}{5}\)