Question

Tìm x

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

Answer 1

= (2x-5)(2x+5)-(2x-5)(2x+7)=0

= (2x-5)(2x+5-2x-7)=(2x-5).(-2)=0

=>2x-5=0=>x=\(\dfrac{5}{2}\)

Answer 2

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow-2\left(2x-5\right)=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy...

Answer 3

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow-2\left(2x-5\right)=0\)

\(\Leftrightarrow2x-5=0\Leftrightarrow x=\dfrac{5}{2}\)

Vậy x = 5/2