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để M xác định
\(\Rightarrow\orbr{\begin{cases}y-1\ne0\\y+1\ne0\end{cases}}\Rightarrow\frac{y\ne1}{y\ne-1}.\)
\(b,M=\frac{1}{y-1}+\frac{y}{y+1}+\frac{2y^2}{y^2-1}\)
\(M=\frac{y+1}{\left(y+1\right)\left(y-1\right)}+\frac{y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)}+\frac{2y^2}{\left(y+1\right)\left(y-1\right)}\)
\(M=\frac{y+1-y^2+y+2y^2}{\left(y+1\right)\left(y-1\right)}=\frac{1+2y+y^2}{\left(y+1\right)\left(y-1\right)}=\frac{\left(1+y\right)^2}{\left(y+1\right)\left(y-1\right)}\)
\(M=\frac{y+1}{y-1}\)
c, Để M nhận giá trị nguyên
\(\Rightarrow y+1⋮y-1\)
\(\Leftrightarrow y-1+2⋮y-1\)
\(\Rightarrow y-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
y = .... Tự tính
\(\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right].\frac{4x^2-4}{5}\) \(ĐKXĐ:x\ne\pm1;\)
\(=\)\(\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x^2-1\right)}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right]\)\(.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\frac{x^2+2x+1+6-\left(x^2+2x-3\right)}{2\left(x-1\right)\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
a: ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
\(a,ĐK:x\ne-3;x\ne0;y\ne0\\ b,A=\dfrac{1}{x^2\left(x+3\right)+y^2\left(x+3\right)}=\dfrac{1}{\left(x^2+y^2\right)\left(x+3\right)}\\ x=y=0\Leftrightarrow A\in\varnothing\)
a) y ≠ − 5 ; 1 2 b) y ≠ 0 ; 5