ĐK: ` x \ne 10; x \ne 0`

`120/(x-10)-3/5=120/x`

`<=>120/(x-10)-120/x=3/5`

`<=>1/(x-10) - 1/x= 1/200`

`<=> (x-x+10)/(x(x-10)) = 1/200`

`<=> 10/(x(x-10))= 1/200`

`<=> x^2-10=2000`

`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)

Vậy `S={50;-40}`.

`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`

`<=>40/(x-10)-1/5=40/x`

`<=>200x-x(x-10)=200(x-10)`

`<=>200x-200x+2000-x^2+10x=0`

`<=>x^2-10x-2000=0`

`Delta'=25+2000=2025`

`<=>x_1=50,x_2=-40`

Vậy `S={50,-40}`

Đặt \(\hept{\begin{cases}\sqrt{7x+11}=a\\\sqrt{9-7x}=b\end{cases}}\)

\(\Rightarrow a^2-b^2=14x+2\)

\(\Rightarrow\frac{2}{a^2-b^2}+\frac{1}{ab}=\frac{7}{24}\)

\(\Leftrightarrow\left(b+7a\right)\left(7b-a\right)=0\)

Làm nhầm phần phân tích nhân tử giờ làm lại cách khác.

Đặt \(7x+11=a\)

\(\Rightarrow7x=a-11\)

\(\Rightarrow\frac{1}{a-10}+\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}\)

\(\Leftrightarrow\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}-\frac{1}{a-10}\)

\(\Leftrightarrow\frac{1}{a\left(20-a\right)}=\left(\frac{7}{24}-\frac{1}{a-10}\right)^2\)

\(\Leftrightarrow\left(a-18\right)\left(a-16\right)\left(49a^2-630a+200\right)=0\)

PS: Bài giải trên bỏ đi nha

\(\frac{96}{x-4}-\frac{120}{x+4}=1\)

\(\Leftrightarrow\frac{96}{x-4}-\frac{120}{x+4}-1=0\)

\(\Leftrightarrow\frac{96\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{120\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Leftrightarrow\frac{96x+384}{x^2-2^2}-\frac{120x-480}{x^2-2^2}-\frac{x^2-2^2}{x^2-2^2}=0\)

\(\Leftrightarrow\left(96x+384-120x+480-x^2+2^2\right)\cdot\frac{1}{x^2-2^2}=0\)

\(\Leftrightarrow-x^2-24x+868=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7x}{x-6}+\dfrac{4}{y+10}=3\\\dfrac{-x}{x-6}+\dfrac{3y}{y+10}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7x+42-42}{x-6}+\dfrac{4}{y+10}=3\\\dfrac{-x+6-6}{x-6}+\dfrac{3y+30-30}{y+10}=-11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{42}{x-6}-\dfrac{4}{y+10}=-10\\\dfrac{-6}{x-6}+\dfrac{-30}{y+10}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{42}{x-6}-\dfrac{4}{y+10}=-10\\\dfrac{-42}{x-6}+\dfrac{-210}{y+10}=-70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-214}{y+10}=-80\\\dfrac{-6}{x-6}+\dfrac{-30}{y+10}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-293}{40}\\x=\dfrac{69}{65}\end{matrix}\right.\)

\(pt\Leftrightarrow2x\sqrt{10+x}-2x\sqrt{10-x}-3\sqrt{100-x^2}=0\)

\(+\text{TH1: }x>0\)

\(pt\Leftrightarrow2x\left(\sqrt{10+x}-4\right)+2x\left(4-\sqrt{10-x}\right)+3\left(8-\sqrt{100-x^2}\right)=0\)

\(\Leftrightarrow2x.\frac{10+x-4^2}{\sqrt{10+x}+4}+2x.\frac{2^2-10+x}{\sqrt{10-x}+2}+3.\frac{8^2-100+x^2}{8+\sqrt{100-x^2}}=0\)

\(\Leftrightarrow\left(x-6\right)\left[\frac{2x}{\sqrt{10+x}+4}+\frac{2x}{\sqrt{10-x}+2}+\frac{3\left(x+6\right)}{\sqrt{100-x^2}+8}\right]=0\)

\(\Leftrightarrow x=6.\)

Đặt \(t=-x\Rightarrow t>0\)

Và \(2t\sqrt{10+t}-2t\sqrt{10-t}-3\sqrt{100-t^2}=0\)

\(+\text{Nếu }t>6\text{ thì }VT>0\)

\(+\text{Nếu }t<6\)\(\text{thì }VT<0\)

\(+\text{Nếu }t=6\text{ thì }VT=0=VP\)

Vậy \(t=6\Rightarrow x=-6\)

Cách 2 đánh giá có vẻ nhanh.

ĐKXĐ: x ≠ \(\pm\) 1

Từ phương trình ban đầu suy ra:

\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)

⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)

⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)

⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)

⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)

Vậy ...