Question

giải phương trình

\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

Answer 1

\(\Leftrightarrow\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)

\(\Leftrightarrow8\left(x-2\right)^2-3\left(2x-3\right)\left(x+3\right)+4\left(x-4\right)^2=0\)

\(\Leftrightarrow8\left(x^2-4x+4\right)-3\left(2x^2+3x-9\right)+4\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-6x^2-9x+27+4x^2-32x+64=0\)

\(\Leftrightarrow6x^2-73x=-123\)

\(\Leftrightarrow x\left(6x-73\right)=-123\)

\(\Leftrightarrow x=-123\) hay \(6x=-50\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-123\\x=-\frac{50}{6}\end{matrix}\right.\)

Answer 2

\(\frac{8}{ }\)