\(\frac{9x}{2x^2+x+3}-\frac{x}{2x^2-x+3}=8\)

\(\Leftrightarrow9x\left(2x^2-x+3\right)-x\left(2x^2+x+3\right)=8\left(2x^2+x+3\right)\left(2x^2-x+3\right)\)

\(\Leftrightarrow16x^3-10x^2+35x=32x^4-88x^2+88x-192\)

\(\Leftrightarrow16x^3-10x^2+35x-32x^4+88x^2-88x+192=0\)

\(\Leftrightarrow16x^3+78x^2-53x-32x^4+192=0\)

Nhưng vì \(16x^3+78x^2-53x-32x^4+192\ne0\)

Nên: phương trình vô nghiệm.

Đặt \(\hept{\begin{cases}\sqrt{x-1}=a\\\sqrt{x+1}=b\end{cases}\left(a;b>0\right)\Rightarrow}b^2-a^2=0\)

\(b^2+2+ab=3b+a\)

\(\Leftrightarrow a\left(b-1\right)+\left(b^2-2b+1\right)-\left(b-1\right)=0\)

\(\Leftrightarrow a\left(b-1\right)+\left(b-1\right)^2-\left(b-1\right)=0\)

\(\Leftrightarrow\left(b-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=1\\a+b=2\end{cases}}\)

Tự làm nốt nhé~~~

\(\sqrt{98}-\sqrt{72}-0,5\sqrt{8}.\)

\(=\sqrt{7\cdot7\cdot2}-\sqrt{6\cdot6\cdot2}-0,5\sqrt{2\cdot2\cdot2}\)

\(=7\sqrt{2}-6\sqrt{2}-0,5\cdot2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}-\sqrt{2}\)

\(=0\)

\(\sqrt{98}-\sqrt{72}-\frac{\sqrt{8}}{2}\)

=\(7\sqrt{2}-6\sqrt{2}-\sqrt{2}\)

\(=0\)

\(D=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)

   \(=\left(2\sqrt{3}-1\right)^2\left(\sqrt{3}+2\right)^2-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)

   \(=\left(4+3\sqrt{3}\right)^2-8\sqrt{28+6\sqrt{3}}\)\(=\left(4+3\sqrt{3}\right)^2-8\left(3\sqrt{3}+1\right)\)

   \(=43+24\sqrt{3}-24\sqrt{3}-8=35\)

\(a^2+b^2=\frac{9a^2}{9}+\frac{16b^2}{16}\ge\frac{\left(3a+4b\right)^2}{9+16}=\frac{5^2}{25}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{3a}{9}=\frac{4b}{16}=\frac{3a+4b}{9+16}=\frac{5}{25}=\frac{1}{5}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=\frac{3}{5}\\b=\frac{4}{5}\end{cases}}\)