a) \(\frac{x-1}{x}+\frac{1-x}{x^2+x}=\frac{1}{x+1}\)
ĐKXĐ : \(x\ne0;x\ne-1\)
Với điều kiện trên, ta có :
\(\Leftrightarrow\frac{x^2-1}{x\left(x+1\right)}+\frac{1-x}{x\left(x+1\right)}=\frac{x}{x\left(x+1\right)}\)
\(\Leftrightarrow x^2-1+1-x=x\)
\(\Leftrightarrow x^2-x-x=1-1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(khôngthỏamãn\right)\\x=2\left(thõamãn\right)\end{matrix}\right.\)
Vậy : \(S=\left\{2\right\}\)
c) \(\frac{x}{6}-97-\frac{x}{12}=5\)
\(\Leftrightarrow2x-1164-x=60\)
\(\Leftrightarrow x=1224\)
Vậy : \(S=\left\{1224\right\}\)
/đề bạn ghi không rõ lắm nên mình làm như thế này nhé
a) \(\frac{X-1}{X+1}\) - \(\frac{x}{x^2+x}\) = \(\frac{1}{x+1}\) ⇌ x2 - x - x = x ⇌ x2 - x - x - x = 0
⇌ x2 - 3x = 0 ⇌ x(x - 3) =0
⇒\(\left\{{}\begin{matrix}x=0\\x-3=0\rightarrow x=3\end{matrix}\right.\)
→ S = \(\left\{0;3\right\}\)
b, | x - 1 |- x = 0 ⇌ | x - 1 |
⇒\(\left[{}\begin{matrix}x-1khix-1\ge0hayx\ge1\\-\left(x-1\right)=1-xkhix-1\le0hayx< 1\end{matrix}\right.\)
+ TH1 : nếu x - 1 \(\ge0hayx\ge1\)
⇒ | x - 1 | - x = 0 ⇌ x - 1 - x = 0 ⇌ -1 = 0 ( vô lý ) (L)
+ TH2 : nếu \(x-1< 0hayx< 1\)
⇒ | x - 1 | - x = 0 ⇌ 1 - x - x = 0 ⇌ -2x = -1 ⇌ x = \(\frac{1}{2}\) ( tmđk)
→ S = \(\left\{\frac{1}{2}\right\}\)
c, \(\frac{x}{6}\) - 97 - \(\frac{x}{12}\) = 5 ⇌ 2x - 1164 - x = 60 ⇌ 2x - x = 60 + 1164 ⇌ x=1224
→ S = \(\left\{1224\right\}\)
