đặt \(2x^2+x+9=a\)

\(2x^2-x+1=b\)

=>\(\sqrt{a}+\sqrt{b}=\frac{a-b}{2}\)

=>\(2\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}-2\right)=0\)

=>\(\sqrt{a}-\sqrt{b}-2=0\)

=>\(a=b+4\sqrt{b}+4\)

=>\(2x^2+x+9=2x^2-x+1+4\sqrt{2x^2-x+1}+4\)

=>\(2x+4=\sqrt{2x^2-x+1}\)

đến đây tự giải nốt

ĐKXĐ: \(\hept{\begin{cases}\sqrt{x}\ne2\\x\ne4\end{cases}\Rightarrow x\ne4}\)

\(P=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\right]:\frac{2.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\)

\(P=\frac{2\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\cdot\frac{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}{2\sqrt{x}+4}=\frac{2\sqrt{x}-2}{2\sqrt{x}+4}\)

\(\frac{3x+1}{2}-x=1\)

\(\Leftrightarrow\frac{3x+1}{2}-\frac{2x}{2}=\frac{2}{2}\)

\(\Rightarrow3x+1-2x=2\)

\(\Leftrightarrow3x-2x=2-1\)

\(\Leftrightarrow x=1\)

Vậy tập nghiệm của pt là S={1}

\(A=x+y+\frac{1}{2x}+\frac{2}{y}\)

\(A=\frac{x}{2}+\frac{1}{2x}+\frac{y}{2}+\frac{2}{y}+\frac{x}{2}+\frac{y}{2}\)

\(A=\left(\frac{x}{2}+\frac{1}{2x}\right)+\left(\frac{y}{2}+\frac{2}{y}\right)+\left(\frac{x}{2}+\frac{y}{2}\right)\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{2y}{2y}}+\frac{3}{2}=1+2+\frac{3}{2}=\frac{9}{2}\)

\("="\Leftrightarrow x=1;y=2\)

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

\(\Leftrightarrow y=\frac{x^2}{2}\)

Thê vô  (2) được

\(2x^2+\left(\frac{x^2}{2}\right)^2+x=14\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2+12x+28\right)=0\)

cảm ơn alibaba =))