ĐKXĐ: \(x>0\)

Ta có:

\(-\sqrt{x}-2\left(x-\frac{1}{x}\right)=\frac{1}{2x^3}-\frac{1}{2x\sqrt{x}}\)

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

\(\frac{\Leftrightarrow1}{2x\sqrt{x}}-\sqrt{x}=2\left(x-\frac{1}{x}+\frac{1}{4x^3}\right)\)

Đặt : \(\frac{1}{2x\sqrt{x}}-\sqrt{x}=a\Rightarrow a^2=x-\frac{1}{x}+\frac{1}{4x^3}\)

Khi đó pt đã cho trở thành:

\(a=2a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{1}{2}\end{cases}}\)

+) a = 0\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Tương tự

ĐKXĐ: ...

\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}+\left(2x+5\right)^2=8\)

\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}-2.\frac{3\left(2x+5\right)}{2\left(x+4\right)}.\left(2x+5\right)+\left(2x+5\right)^2+\frac{3\left(2x+5\right)^2}{x+4}=8\)

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

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

Đặt \(\frac{\left(2x+5\right)^2}{x+4}=a\)

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

Nghiệm xấu, bạn tự giải nốt

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}\\ \Leftrightarrow\frac{2x^2-8x}{\left(x+1\right)\cdot\left(x-4\right)}-\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{2x^2-8x-x^2+x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{x^2-7x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow x^2-7x-8=0\\ \Rightarrow\left[{}\begin{matrix}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\)

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\) (ĐKXĐ: x \(\ne\) 1)

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

\(\Leftrightarrow\) 2x(x + 1) = x² - x + 8

\(\Leftrightarrow\) 2x² + 2x - x² + x - 8 = 0

\(\Leftrightarrow\) x² + 3x - 8 = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\))² - \(\frac{41}{4}\) = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\) - \(\frac{\sqrt{41}}{2}\))(x + \(\frac{3}{2}\) + \(\frac{\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\) (x + \(\frac{3-\sqrt{41}}{2}\))(x + \(\frac{3+\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3-\sqrt{41}}{2}=0\\x+\frac{3+\sqrt{41}}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{41}-3}{2}\\x=\frac{-3-\sqrt{41}}{2}\end{matrix}\right.\)

Vậy S = {\(\frac{\sqrt{41}-3}{2}\); \(\frac{-3-\sqrt{41}}{2}\)}

Chúc bn học tốt!!

Bạn xem lại xem có viết nhầm đề bài không thế?

Bài này được giải trên onl math rồi

Đặt \(\sqrt{x}=t\left(t>0\right)\)

\(\Leftrightarrow\frac{1}{1+t^2}+\frac{2}{1+t}=\frac{2+t}{2t^2}\)

\(\Leftrightarrow\frac{1+t+2t+2t^2}{\left(1+t\right)\left(1+t^2\right)}=\frac{2+t}{2t^2}\)

\(\Leftrightarrow\frac{2t^2+3t+1}{\left(1+t\right)\left(1+t^2\right)}=\frac{2+t}{2t^2}\)

\(\Leftrightarrow\frac{\left(t+1\right)\left(2t+1\right)}{\left(1+t\right)\left(1+t^2\right)}=\frac{2+t}{2t^2}\)

\(\Leftrightarrow\frac{2t+1}{1+t^2}=\frac{2+t}{2t^2}\)

\(\Leftrightarrow2t^2\left(2t+1\right)=\left(2-t\right)\left(1+t^2\right)\)

\(\Leftrightarrow4t^3+2t^2=2+2t^2+1+t^3\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1\)

\(x=0\) không phải nghiệm, pt tương đương:

\(\frac{12}{x+4+\frac{2}{x}}-\frac{3}{x+2+\frac{2}{x}}=1\)

Đặt \(x+2+\frac{2}{x}=a\)

\(\frac{12}{a+2}-\frac{3}{a}=1\Leftrightarrow12a-3\left(a+2\right)=a\left(a+2\right)\)

\(\Leftrightarrow a^2-7a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2+\frac{2}{x}=1\\x+2+\frac{2}{x}=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-4x+2=0\end{matrix}\right.\)