a) A có nghĩa khi \(\hept{2x-2\ne02-2x^2\ne0\Leftrightarrow\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}\Leftrightarrow}x\ne\pm1}\)
Vậy A có nghĩa khi \(x\ne\pm1\)
b) \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow A=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
Vậy A=\(\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)
b) \(A=\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)
A=\(\frac{-1}{2}\)\(\Leftrightarrow\frac{1}{2\left(x-1\right)}=\frac{-1}{2}\)
\(\Leftrightarrow-2\left(x-1\right)=2\)
<=> x-1=-1
<=> x=0 (tmđk)
Vậy x=0 thì \(A=\frac{-1}{2}\)
a) \(x\ne1,2;x\inℝ\)
a) Để biểu thức A có nghĩa \(\hept{\begin{cases}2-2x^2\ne0\\2x-2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}2\left(1-x\right)\left(1+x\right)\ne0\\2\left(x-1\right)\ne0\end{cases}}\Rightarrow x\ne\pm1\)
b) Rút gọn \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x+1\right)}\)
c) \(A=-\frac{1}{2}\Leftrightarrow\frac{1}{2\left(x+1\right)}=-\frac{1}{2}\Leftrightarrow\frac{1}{x+1}=-1\Leftrightarrow x+1=-1\Leftrightarrow x=-2\)Vậy x=-2 Thì A=-1/2
Khánh Hoàng Câu b bạn làm sai rồi nhé. Bạn chưa đổi dấu phải là -1
a) ĐK : \(\hept{\begin{cases}2.x-2\ne0\\2-2.x^2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)
b) A = \(\frac{x}{2.x-2}+\frac{x^2+1}{2-2.x^2}\)
\(=\frac{x}{2.\left(x-1\right)}+\frac{x^2+1}{-2.\left(x^2-1\right)}\)
\(=\frac{x}{2.\left(x-1\right)}-\frac{x^2+1}{2.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x.\left(x+1\right)-\left(x^2+1\right)}{2.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x^2+x-x^2-1}{2.\left(c-1\right).\left(x+1\right)}\)
\(=\frac{x-1}{2.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{1}{2.\left(x+1\right)}\)
c) A=\(-\frac{1}{2}\)<=> \(\frac{1}{2.\left(x+1\right)}=-\frac{1}{2}\)
<=> x + 1 = -2
<=> x = -2
Vậy x=- 2
