Question

Tìm x

a)\(\left(x+4\right)^2+\left(2x-5\right)^2=5x^2+19\)

b) \(\left(2x-5\right)^2-\left(2x-6\right)^2=7x-12\)

c) \(\left(3x+10\right)^2-\left(2x-7\right)^2=0\)

d) \(\left(5x+12\right)-\left(3x-9\right)^2=0\)

e) \(\left(4x-3\right)^2-\left(2x-7\right)^2=12x^2+81\)

Answer 1

\(x^2+8x+16+4x^2-20x+25-5x^2-19=0.\\ \Leftrightarrow-12x-22=0.\\ \Leftrightarrow-12x=22.\\ \Leftrightarrow x=\dfrac{-11}{6}.\)

\(\Leftrightarrow4x^2-20x+2-4x^2+24x-36-7x+12=0.\\ \Leftrightarrow-3x-22=0.\\ \Leftrightarrow x=\dfrac{22}{3}.\)

\(\Leftrightarrow\left(3x+10-2x+7\right)\left(3x+10+2x-7\right)=0.\\ \Leftrightarrow\left(x+17\right)\left(5x+3\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=-\dfrac{3}{5}.\end{matrix}\right.\)