ĐK: \(x\ne\pm2\)

Phương trình đã cho tương đương với: \(\left(\frac{x+3}{x-2}\right)^2+6\left(\frac{x-3}{x+2}\right)^2-7\left(\frac{x+3}{x-2}.\frac{x-3}{x+2}\right)=0\)(1)

Đặt \(\frac{x+3}{x-2}=t,\frac{x-3}{x+2}=k\)

Khi đó (1) trở thành: \(t^2+6k^2-7tk=0\)

\(\Leftrightarrow t\left(t-6k\right)-k\left(t-6k\right)=0\Leftrightarrow\left(t-k\right)\left(t-6k\right)=0\Leftrightarrow\orbr{\begin{cases}t=k\\t=6k\end{cases}}\)

- Nếu t = k thì \(\frac{x+3}{x-2}=\frac{x-3}{x+2}\Rightarrow\left(x+3\right)\left(x+2\right)=\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow x^2+5x+6=x^2-5x+6\Rightarrow5x=-5x\Rightarrow x=0\)(thỏa mãn điều kiện)

- Nếu t = 6k thì \(\frac{x+3}{x-2}=6.\frac{x-3}{x+2}\) 

\(\Rightarrow\left(x+3\right)\left(x+2\right)=6\left(x-3\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+5x+6=6x^2-30x+36\)

\(\Leftrightarrow6x^2-30x+36-x^2-5x-6=0\)

\(\Leftrightarrow5x^2-35x+30=0\Leftrightarrow5\left(x^2-7x+6\right)=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\) (thỏa mãn điều kiện)

Vậy tập nghiệm của phương trình là \(S=\left\{0;1;6\right\}\)

Ta có: \(x^2+\frac{1}{x^2}-\frac{9}{2}\left(x+\frac{1}{x}\right)+7=0\)

<=> \(\left(x^2+\frac{1}{x^2}+2\right)-\frac{9}{2}\left(x+\frac{1}{x}\right)+5=0\)

,<=> \(\left(x+\frac{1}{x}\right)^2-\frac{9}{2}\left(x+\frac{1}{x}\right)+\frac{81}{16}-\frac{1}{16}=0\)

<=> \(\left(x+\frac{1}{x}-\frac{9}{4}\right)^2=\frac{1}{16}\)

<=> \(\orbr{\begin{cases}x+\frac{1}{x}-\frac{9}{4}=\frac{1}{4}\\x+\frac{1}{x}-\frac{9}{4}=-\frac{1}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x+\frac{1}{x}-\frac{5}{2}=0\\x+\frac{1}{x}-2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2-\frac{5}{2}x+1=0\\x^2-2x+1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)=\frac{9}{16}\\\left(x-1\right)^2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-\frac{5}{4}\right)^2=\frac{9}{16}\\x-1=0\end{cases}}\)

<=> x - 5/4 = 3/4 hoặc x - 5/4 = -3/4

 hoặc x = 1

<=> x = 2 hoặc x = 1/2

hoặc x = 1

Vậy S = {1/2; 1; 2}

Đáp án:

1/2;1;2

bye

a/ Đặt \(\hept{\begin{cases}\frac{x+1}{x-2}=a\\\frac{x+1}{x-4}=b\end{cases}}\) thì có

\(a^2+b-\frac{12b^2}{a^2}=0\)

\(\Leftrightarrow\left(a^2-3b\right)\left(a^2+4b\right)=0\)

b/ \(2x^2+3xy-2y^2=7\)

\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)

ĐKXĐ: ....

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(t^2-2+\frac{9}{2}t+7=0\)

\(\Leftrightarrow2t^2+9t+10=0\Rightarrow\left[{}\begin{matrix}t=-2\\t=-\frac{5}{2}\end{matrix}\right.\)

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