ĐKXĐ: ...

\(\Leftrightarrow\frac{2\left(x^2+1\right)}{\left(1-x^2\right)^2}+\frac{1}{4x^2}=\frac{\left(3x^2+1\right)^2}{144}\)

Đặt \(\left\{{}\begin{matrix}1-x^2=a\\4x^2=b\end{matrix}\right.\)

\(\Rightarrow\frac{2a+b}{a^2}+\frac{1}{b}=\frac{\left(a+b\right)^2}{144}\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{a^2b}=\frac{\left(a+b\right)^2}{144}\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\left(vn\right)\\a^2b=144\end{matrix}\right.\)

\(\Leftrightarrow\left(1-x^2\right)^2.4x^2=144\)

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

\(\Leftrightarrow...\)

=> 144x - 100(x + 2) = 2x(x + 2)

<=> 144x - 100x - 200 =  2x² + 4x

 <=> - 2x²  + 40x - 200 = 0

<=> -2x²  + 20x + 20x - 200 = 0

<=> (x - 10)(-2x + 20) = 0

<=> x - 10 = 0 hoặc -2x + 20 = 0 

<=> x = 10 

 cảm ơn câu hỏi của bạn nka !!!    ĐKXĐ của phương trình \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

\(\frac{144}{x+2}-\frac{100}{x}=2\)

\(\Rightarrow\frac{144x}{x\left(x+2\right)}-\frac{100\left(x+2\right)}{x\left(x+2\right)}=\frac{2x\left(x+2\right)}{x\left(x+2\right)}\) 

\(\Rightarrow144x-100x-200=2x^2+4x\) 

 \(\Leftrightarrow144x-100x-4x-200-2x^2=0\)

\(\Leftrightarrow40x-200-2x^2=0\)

 \(\Leftrightarrow-2\left(x^2-20x+100\right)=0\)

 \(\Leftrightarrow-2\left(x-10\right)^2=0\)\(\Leftrightarrow\left(x-10\right)^2=0\)\(\Leftrightarrow x=10\)(NHẬN)

 vậy tập nghiệm của phương trình là   S= 10

Theo bài ra , ta có :

\(\frac{144}{x+2}-\frac{100}{x}=2\left(ĐKXĐ:x\ne0;x\ne2\right)\)

Quy đồng cà khử mẫu ta được :

\(144x-100\left(x+2\right)=2x\left(x+2\right)\)

\(\Leftrightarrow144x-100x-200=2x^2+4x\)

\(\Leftrightarrow44x-200-2x^2-4x=0\)

\(\Leftrightarrow-2x^2+40x-200=0\)

\(\Leftrightarrow-2\left(x^2-20x+100\right)=0\)

\(\Leftrightarrow x^2-20x+100=0\)

\(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

Vậy \(S=\left\{10\right\}\)

Chúc bạn hok tốt =))

\(\frac{144}{x+2}-\frac{100}{x}=2\left(1\right)\)

ĐKXĐ : \(x\ne-2;x\ne0\)

MTC : x(x + 2 )

\(\left(1\right)\Leftrightarrow\frac{144x}{x\left(x+2\right)}-\frac{100\left(x+2\right)}{x\left(x+2\right)}=\frac{2x\left(x+2\right)}{x\left(x+2\right)}\)

\(\Leftrightarrow144x-100x-200=2x^x+4x\)

\(\Leftrightarrow2x^2+4x-144x+100x+200=0\)

\(\Leftrightarrow2x^2-40x+200=0\)\(\Leftrightarrow2\left(x^2-20x+10^2\right)=0\)

\(\Leftrightarrow2\left(x-10\right)^2=0\)\(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Leftrightarrow x-10=0\Leftrightarrow x=10\left(chọn\right)\)

Vậy tập nghiệm của phương trình là S = { 10 }

\(\dfrac{144}{x+2}\)\(-\)\(\dfrac{100}{x}\)=2

\(\Leftrightarrow\)\(\dfrac{144x}{\left(x+2\right)x}-\dfrac{100\left(x+2\right)}{x\left(x+2\right)}=\dfrac{2x\left(x+2\right)}{x\left(x+2\right)}\)

\(\Leftrightarrow144x-100\left(x+2\right)=2x\left(x+2\right)\)

\(\Leftrightarrow144x-100x-200=2x^2+4x\)

\(\Leftrightarrow-2x^2+144x-100x-4x=200\)

\(\Leftrightarrow-2x^2+40x=200\)

\(\Leftrightarrow-2x^2+40x-200=0\)

\(\Leftrightarrow-2\left(x^2-20x+100\right)=0\)

\(\Leftrightarrow x^2-20x+100=0\)

\(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

mình làm từng bước mong bạn sẽ hiểu !!!

CHÚC BẠN HC TỐT NHA !!!

\(\Leftrightarrow\frac{200\left(x+20\right)}{2x\left(x+20\right)}-\frac{240x}{2x\left(x+20\right)}=\frac{x\left(x+20\right)}{2x\left(x+20\right)}\) đk: x\(\ne0\) , x \(\ne-20\)

\(\Rightarrow200x+4000-240x=x^2+20x\)

\(\Leftrightarrow-x^2-60x+4000=0\)

\(\Leftrightarrow x^2+60x-4000=0\)

\(\Leftrightarrow x^2+100x-40x-4000=0\)

\(\Leftrightarrow\left(x+100\right)\left(x-40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+100=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-100\left(tmđk\right)\\x=40\left(tmđk\right)\end{matrix}\right.\)

Vậy S\(=\left\{-100;40\right\}\)

\(\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2}\)

\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2},x\ne0,x\ne-20\)

\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{200\left(x+20\right)-240x-x\left(x+20\right)}{2x\left(x+20\right)}=0\)

\(\Leftrightarrow\frac{200x+4000-240x-x^2-20x}{2x\left(x+20\right)}=0\)

\(\Leftrightarrow-60x+4000-x^2=0\)

\(\Leftrightarrow-x^2-60x+4000=0\)

\(\Leftrightarrow x^2+60x-4000=0\)

\(\Leftrightarrow\frac{-60\pm\sqrt{60^2}-4.1\left(-4000\right)}{2}\)

\(\Leftrightarrow\frac{-60\pm\sqrt{3600+16000}}{2}\)

\(\Leftrightarrow\frac{-60\pm\sqrt{19600}}{2}\)

\(\Leftrightarrow\frac{-60\pm140}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{-60+140}{2}\\\frac{-60-140}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-100\end{matrix}\right.,x\ne0,x\ne-20\)

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

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\left(x\ne0;x\ne2\right)\)

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

\(\Rightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)

Vậy pt có nghiệm x =-1

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\\ \frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x^2-2x}=0\\ \frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}-\frac{2}{x\left(x-2\right)}=0\\ x^2+2x-x+2-2=0\\ x\left(x+1\right)=0\\ x=0,x=-1\)

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

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x};x\ne2;x\ne0\)

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

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-\left(x-2\right)-2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-x+2-2}{x\times\left(x-2\right)}=0\)

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

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

\(\Leftrightarrow\frac{x+1}{x-2}=0\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~

\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)

Mấy bài kia sao cái phương trình dài thê,s giải sao nổi

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\Rightarrow x=1\\\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}=0\left(1\right)\end{matrix}\right.\)

Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:

\(\frac{2}{a+1}-\frac{1}{a-1}-\frac{1}{a}=0\Leftrightarrow\frac{2\left(a-1\right)-\left(a+1\right)}{\left(a^2-1\right)}-\frac{1}{a}=0\)

\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)

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

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

Vậy \(x=1\)