a) Biểu thức M xác định <=> \(\hept{\begin{cases}2-2x\ne0\\2-2x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x^2\ne1\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}}\)
Vậy đk xác định biểu thức M <=> x \(\ne\)\(\pm\)1
b) Ta có:
M = \(\frac{x}{2-2x}-\frac{x^2+1}{2-2x^2}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x^2\right)}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x\left(x+1\right)}{2\left(1-x\right)\left(x+1\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x^2+x-x^2-1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x-1}{-2\left(x-1\right)\left(x+1\right)}\)
M = \(-\frac{1}{2\left(x+1\right)}\) (đk : x + 1 \(\ne\)0 => x \(\ne\)-1)
