Giải:
\(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(5x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(25x^2+20x+4\right)-\left(x^2-4x-5\right)+x^2+8x+16\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=125x^2+100x+20-x^2+4x+5+x^2+8x+16\)
\(\Leftrightarrow5x^2+12x+8=125x^2+112x+41\)
\(\Leftrightarrow125x^2+112x+41-5x^2-12x-8=0\)
\(\Leftrightarrow120x^2+100x+33=0\) (vô lí)
Vậy phương trình vô nghiệm
Đúng 0
Bình luận (0)
