\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)

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

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

\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)

<=>(2-x)(2+x)=0

<=>x=2 hoặc x=-2

Vậy S={-2;2}

mk làm rồi đó bạn xem đi Xem câu hỏi

hau ta cai j ?

11x^2-490x-3000=0

<=> 11x^2+60x-550x-3000=0

<=> 11x(x-50)-60(x-50)=0

<=> (x-50)(11x-60)=0

<=> x=50 hoặc x=60/11

\(4x^4-11x^2+6=0\)

\(\Leftrightarrow4x^4-8x^2-3x^2+6=0\)

\(\Leftrightarrow4x^2\left(x^2-2\right)-3\left(x^2-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-2=0\\4x^2-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{2}\\x=\pm\sqrt{\frac{3}{4}}\end{cases}}\)

\(S=\left\{\pm\sqrt{2};\pm\sqrt{\frac{3}{4}}\right\}\)

nếu có sai  bn thông cảm nha