Question

\(\text{b)(2x+3)^2}+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)

đề:tìm x

Answer 1

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

\(\Leftrightarrow\left(2x+3\right)^2-\left(x+4\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-5\right)\left(x+1\right)=5\left(x^2+4x+4\right)\)

\(\Leftrightarrow\left(2x+3+x+4\right)\left(2x+3-x-4\right)+\left(x+1\right)\left(x-1+x-5\right)=5x^2+20x+20\)

\(\Leftrightarrow\left(3x+7\right)\left(x-1\right)+\left(x+1\right)\left(2x-6\right)=5x^2+20x+20\)

\(\Leftrightarrow3x^2-3x+7x-7+2x^2-6x+2x-6=5x^2+20x+20\)

\(\Leftrightarrow5x^2-13-5x^2-20=20x\)

\(\Leftrightarrow-33=20x\)

\(\Leftrightarrow x=\dfrac{-33}{20}\)

\(\Rightarrow S=\left\{\dfrac{-33}{20}\right\}\)