3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)
Mc: \(x.\left(x-5\right)\)
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0
\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0
\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3
Vậy \(x\) = 0 hoặc \(x\) = 3
\(x-5\ne0\Rightarrow x\ne5\)
\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {3}
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
Mc: \(x.\left(x+7\right)\)
\(\Leftrightarrow x^2-4x-x-7=-7\)
\(\Leftrightarrow x^2-4x-x=-7+7\)
\(\Leftrightarrow\) \(x^2-5x=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x-5=0\)
\(\Leftrightarrow x=0\) hoặc \(x=5\)
Vậy \(x=0\) hoặc \(x=5\)
\(x+7\ne0\Rightarrow x\ne-7\)
\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {5}
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Mc : \(\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)
\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)
\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=2\) hoặc \(x=2\)
\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
MC: \(\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)
\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4.\left(x-1\right)=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)
\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(Mc:\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)
\(\Leftrightarrow0=0\) (Nhận)
Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)
sai ngay dòng đầu
\(x^2=0\Leftrightarrow\frac{x}{a}=\frac{0}{x}\)
vì khi x2=0 <=> x=0, mà x nằm ở mẫu thức => vô lí
Sai ở ngay đầu dòng :
Do x2 = 0 => x = 0
Mà x nằm ở mẫu -> Vô lý.