Question

(2 x + 3)( 2x + 1) − (2x + 5)( 2 x + 7) = 1 −( 6 x² + 9 x − 9)

Answer 1

Ta có: \(\left(2x+3\right)\left(2x+1\right)-\left(2x+5\right)\left(2x+7\right)=1-\left(6x^2+9x-9\right)\)

\(\Leftrightarrow4x^2+2x+6x+3-\left(4x^2+14x+10x+35\right)=1-6x^2-9x+9\)

\(\Leftrightarrow4x^2+8x+3-4x^2-24x-35-1+6x^2+9x-9=0\)

\(\Leftrightarrow6x^2-7x-42=0\)

\(\Delta=49-4\cdot6\cdot\left(-42\right)=1057\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{1057}}{12}\\x_2=\dfrac{7+\sqrt{1057}}{12}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7-\sqrt{1057}}{12};\dfrac{7+\sqrt{1057}}{12}\right\}\)