\(\Rightarrow5-x^2+x^2+3+2\sqrt{\left(5-x^2\right)\left(x^2+3\right)}=16\)

\(\Rightarrow\sqrt{\left(5-x^2\right)\left(3+x^2\right)}=4\)

\(\Rightarrow\left(5-x^2\right)\left(3+x^2\right)=16\)

\(\Rightarrow-x^4+2x^2+15-16=0\)

\(\Rightarrow x^4-2x^2+1=0\)

\(\Rightarrow\left(x^2-1\right)^2=0\)

\(\Rightarrow x^2=1\Rightarrow x=\pm1\)

\(\frac{x^2+5}{\sqrt{x^2+4}}=\frac{x^2+4+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\)

Áp dụng BĐT Cô Si ,ta có:

\(\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\ge2\sqrt{\sqrt{x^2+4}\cdot\frac{1}{\sqrt{x^2+4}}}=2\)

Đặt \(A=\frac{x^2+5}{\sqrt{x^2+4}}\Leftrightarrow A-2=\frac{x^2+5-2\sqrt{x^2+4}}{\sqrt{x^2+4}}\)

\(A-2=\frac{x^2+4-2\sqrt{x^2+4}+1}{\sqrt{x^2+4}}=\frac{\left(\sqrt{x^2+4}-1\right)^2}{\sqrt{x^2+4}}\ge0\)

\(A\ge2\)

\(6x^4+25x^3+12x^2-25x+6=0.\)

\(\Leftrightarrow\left(2x^2+x-2\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)\left(x+3\right)\left(3x-1\right)=0\)

x²  - 1 + 3(6x + 6)=0

ĐKXĐ: \(x\ne0\)

\(\frac{x-1}{x}>\frac{1}{2}\)

\(\frac{x-1}{x}-\frac{1}{2}>0\)

\(\frac{2\left(x-1\right)-x}{2x}>0\)

\(\frac{2x-2-x}{2x}>0\)

\(\frac{x-2}{2x}>0\)

TH1: \(\hept{\begin{cases}x-2>0\\2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x>0\end{cases}\Rightarrow x>2}}\) 

Kết hợp với đkxđ,ta được: \(x>2,x\ne0\)

TH2: \(\hept{\begin{cases}x-2< 0\\2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x< 0\end{cases}\Rightarrow}x< 2}\)

Kết hợp với đkxđ ,ta được: \(x< 2,x\ne0\)

=.= hk tốt!!