Question

(2x+3)²+(x-1)(x+1)=5(x+2)-(x-5)(x+1)+(x+4)

Answer 1

Ta có: \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)-\left(x-5\right)\left(x+1\right)+\left(x+4\right)\)

\(\Leftrightarrow4x^2+12x+9+x^2-1=5x+10-\left(x^2-4x-5\right)+\left(x+4\right)\)

\(\Leftrightarrow5x^2+12x+8=6x+14-x^2+4x+5\)

\(\Leftrightarrow5x^2+12x+8+x^2-10x-19=0\)

\(\Leftrightarrow6x^2+2x-11=0\)

\(\text{Δ}=2^2-4\cdot6\cdot\left(-11\right)=4+264=268>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-2-2\sqrt{67}}{12}=\dfrac{-1-\sqrt{67}}{6}\\x_2=\dfrac{-1+\sqrt{67}}{6}\end{matrix}\right.\)