Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{4}{x-8+\frac{7}{x}}+\frac{5}{x-10+\frac{7}{x}}=-1\)

Đặt \(x-10+\frac{7}{x}=a\)

\(\frac{4}{a+2}+\frac{5}{a}=-1\)

\(\Leftrightarrow4a+5\left(a+2\right)=-a\left(a+2\right)\)

\(\Leftrightarrow a^2+11a+10=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-10+\frac{7}{x}=-1\\x-10+\frac{7}{x}=-10\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+7=0\\x^2+7=0\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow\frac{2x}{3x^2-4x+1}-\frac{7x}{3x^2+2x+1}=6\)

\(\Leftrightarrow\frac{2}{3x-4+\frac{1}{x}}-\frac{7}{3x+2+\frac{1}{x}}=6\)

Đặt \(3x-4+\frac{1}{x}=a\)

\(\frac{2}{a}-\frac{7}{a+6}=6\)

\(\Leftrightarrow2\left(a+6\right)-7a=6a\left(a+6\right)\)

\(\Leftrightarrow6a^2+41a-12=0\)

Nghiệm xấu, bạn coi lại đề

GPT

\(\frac{3}{3x^2-4x+1}+\frac{13}{3x^2+2x+1}=\frac{6}{x}\)

\(\frac{2}{x^2+1}+\frac{4}{x^2+3}+\frac{6}{x^2+5}=3+\frac{x^2-1}{x^2+6}\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+1-\frac{2}{x^2+1}+1-\frac{4}{x^2+3}+1-\frac{6}{x^2+5}=0\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+\frac{x^2-1}{x^2+1}+\frac{x^2-1}{x^2+3}+\frac{x^2-1}{x^2+5}=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{1}{x^2+6}+\frac{1}{x^2+1}+\frac{1}{x^2+3}+\frac{1}{x^2+5}\right)=0\)

\(\Rightarrow x=\pm1\)

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

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

Đặt \(x+2+\frac{2}{x}=a\)

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

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

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

Đặt \(x^2-4x+5=a\)

\(\frac{5}{a}-a+4=0\)

\(\Leftrightarrow-a^2+4a+5=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4x+5=-1\\x^2-4x+5=5\end{matrix}\right.\)

\(x=0\) không phải nghiệm

\(\frac{4}{x+1+\frac{3}{x}}+\frac{5}{x-5+\frac{3}{x}}=-\frac{3}{2}\)

Đặt \(x-5+\frac{3}{x}=a\)

\(\frac{4}{a+6}+\frac{5}{a}=-\frac{3}{2}\)

\(\Leftrightarrow8a+10\left(a+6\right)=-3a\left(a+6\right)\)

\(\Leftrightarrow3a^2+36a+60=0\Rightarrow\left[{}\begin{matrix}a=-2\\a=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-5+\frac{3}{x}=-2\\x-5+\frac{3}{x}=-10\end{matrix}\right.\) \(\Leftrightarrow...\)

ĐKXĐ:...

\(x^2+\frac{36}{x^2}-4\left(x-\frac{6}{x}\right)-17=0\)

Đặt \(x-\frac{6}{x}=a\Rightarrow a^2=x^2+\frac{36}{x^2}-12\Rightarrow x^2+\frac{36}{x^2}=a^2+12\)

\(a^2+12-4a-17=0\)

\(\Leftrightarrow a^2-4a-5=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=-1\\x-\frac{6}{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2-5x-6=0\end{matrix}\right.\)

Thêm 5 vào hai vế suy ra:

\(\left(x^2-4x+5\right)+\frac{10}{x^2-4x+5}=7\)

Đặt \(t=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\). PT trở thành:

\(t+\frac{10}{t}=7\Leftrightarrow\frac{t^2+10}{t}=7\Leftrightarrow t^2-7t+10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=2\end{matrix}\right.\left(C\right)\). Với t = 5 suy ra \(x^2-4x+5=5\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Với t = 2 suy ra \(x^2-4x+5=2\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\).

Vậy tập hợp nghiệm của PT là S = (0;1;3;4)

ĐKXĐ: ...

\(\Leftrightarrow\frac{49}{\left(x-7\right)^2}+1=\frac{25}{x^2}\)

\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+x^2=25\)

\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+2.\frac{7x}{x-7}.x+x^2-\frac{14x^2}{x-7}=25\)

\(\Leftrightarrow\left(\frac{7x}{x-7}+x\right)^2-\frac{14x^2}{x-7}=25\)

\(\Leftrightarrow\left(\frac{x^2}{x-7}\right)^2-\frac{14x^2}{x-7}-25=0\)

Đặt \(\frac{x^2}{x-7}=a\)

\(\Rightarrow a^2-14a-25=0\)

Nghiệm xấu, bạn tự giải tiếp đoạn cuối