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- Điều kiện \(\hept{\begin{cases}x\ne5\\x\ne-5\end{cases}}\)\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{\left(x-5\right)}{2x\left(x+5\right)}=\frac{x+25}{2\left(x+5\right)\left(x-5\right)}\)\(\Leftrightarrow\frac{2\left(x+5\right)^2-\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)\(\Leftrightarrow x^2+30x+25=x^2+25\Leftrightarrow x=0\)
- Điều Kiện : \(x\ne1\)\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)\(\Leftrightarrow x^2+x+1-3x=2x^2-2x\Leftrightarrow x^2=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)so sánh điều kiện có nghiệm phương trình là : \(x=-1\)
\(\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\)tu giai ra de ma
\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)
Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)
ĐKXĐ : \(x\ne0;x\ne\pm5\)
\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)(ko t/m ĐKXĐ)
Vậy phương trình vô nghiệm.
a) \(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)
\(\Rightarrow\dfrac{5\left(x+5\right)}{15}-\dfrac{3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
* Với \(5\left(x+5\right)-3\left(x-3\right)=0\),
Ta có được đẳng thức đúng
=> 5x + 25 - 3x + 9 = 0
=> 2x + 34 = 0
=> 2x = -34
=> x = -17
* Với 5( x+5 ) - 3 (x-3 ) \(\ne\)0, ta có
\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\dfrac{1}{15}=\dfrac{1}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\left(x-3\right)\left(x+5\right)=15\)
\(\Rightarrow x^2+5x-3x-15-15=0\)
\(\Rightarrow x^2+2x-30=0\)
=> \(\left(x+1-\sqrt{31}\right)\left(x+1+\sqrt{31}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{31}\\x=-1-\sqrt{31}\end{matrix}\right.\)
\(a)\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)(ĐKXĐ: \(x\ne3,x\ne-5\))
\(\Leftrightarrow\dfrac{x+5}{3}-\dfrac{x-3}{5}-\dfrac{5}{x-3}+\dfrac{3}{x+5}=0\\ \Leftrightarrow\dfrac{5\left(x-3\right)\left(x+5\right)^2-3\left(x-3\right)^2\left(x+5\right)-75\left(x+5\right)+45\left(x-3\right)}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow\dfrac{2x^3+38x^2+8x-1020}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow2x^3+38x^2+8x-1020=0\\ \Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\left(TM\right)\\x=-1+\sqrt{31}\left(TM\right)\\x=-1-\sqrt{31}\left(TM\right)\end{matrix}\right.\)
Vậy....
ĐK: ...
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)( ko t/m )
d) tương tự, ngại tính lắm
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}-\frac{3x^2}{x^3-1}=\frac{2x\left(x-1\right)}{x^3-1}\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)
