a, ĐKXĐ của M là :
\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ne0\\2\left(1-x\right)\left(1+x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b, \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2x^2-2}\)
\(=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x-2}\)
c, Để giá trị của biểu thức M bằng \(\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
Vậy x = 2 thfi giá trị của M bằng 1/2
a.
M xác định \(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)
b.
\(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\\ =\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x^2-1\right)}\\ =\dfrac{x\left(x+1\right)}{2\left(x^2-1\right)}-\dfrac{x^2+1}{2\left(x^2-1\right)}\\ =\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)
c.
M=1/2
\(\Leftrightarrow\dfrac{1}{2x+2}=\dfrac{1}{2}\\ \Leftrightarrow2x+2=2\\ \Leftrightarrow x=0\)
