Question

\(\frac{x^2+4}{8}+\frac{x^2+3}{7}+\frac{x^2+2}{6}=\frac{x^2+1}{5}+\frac{x^2}{4}+\frac{x^2-1}{3}\)
GIÚP MÌNH VỚI

Answer 1

\(\Leftrightarrow\frac{x^2+4}{8}-1+\frac{x^2+3}{7}-1+\frac{x^2+2}{6}-1=\frac{x^2+1}{5}-1+\frac{x^2}{4}-1+\frac{x^2-1}{3}-1\)

\(\Leftrightarrow\frac{x^2-4}{8}+\frac{x^2-4}{7}+\frac{x^2-4}{6}-\frac{x^2-4}{5}-\frac{x^2-4}{4}-\frac{x^2-4}{3}=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)

\(\Leftrightarrow x^2-4=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)