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a) \(A=\frac{1}{y-1}-\frac{y}{1-y^2}\left(y\ne\pm1\right)\)
\(\Leftrightarrow A=\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}=\frac{y+1}{\left(y-1\right)\left(y+1\right)}+\frac{y}{\left(y-1\right)\left(y+1\right)}=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\)
Thay y=2 (tm) vao A ta co:
\(A=\frac{2\cdot2+1}{\left(2-1\right)\left(2+1\right)}=\frac{5}{3}\)
Vay \(A=\frac{5}{3}\)voi y=2
b) Ta co: \(\hept{\begin{cases}A=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\left(y\ne\pm1\right)\\B=\frac{y^2-y}{2y+1}=\frac{y\left(y-1\right)}{2y+1}\left(y\ne\frac{-1}{2}\right)\end{cases}}\)
\(\Rightarrow M=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\cdot\frac{y\left(y-1\right)}{2y+1}=\frac{\left(2y+1\right)\cdot y\cdot\left(y-1\right)}{\left(y-1\right)\left(y+1\right)\left(2y+1\right)}=\frac{y}{y+1}\)
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
a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}-\frac{y}{\left(1-y\right)\left(1+y+y^2\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y\left(y^2+y+1\right)}{\left(y+1\right)^2\left(y^2+y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y+1\right)^2}\right):\frac{1}{\left(y-1\right)\left(x+1\right)}\)
\(=\left(\frac{\left(y+1\right)^2+y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)^2}\right).\frac{\left(y-1\right)\left(y+1\right)}{1}=\frac{y^2+2y+1+y^2-y}{y+1}=\frac{2y^2+y+1}{y+1}\)
b, Thay y = 1/2 ta có :
\(\frac{2.\left(\frac{1}{2}\right)^2+\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}+\frac{2}{2}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{12}\)
a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(1-3xy\right).\frac{-x-1}{1-3xy}+\frac{x^2}{x+1}\)
\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`
\(=\frac{-\left(x+1\right)^2+x^2}{x+1}\)
\(=\frac{-x^2-2x-1+x^2}{x+1}\)
\(=\frac{-2x-1}{x+1}\)(1)
b) Thay \(x=-3,y=2014\)vào (1) ta được:
\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)
Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014
a)ĐKXĐ:\(y\ne0,y\ne1,y\ne-1\)
b)A\(=\frac{y\left(y-1\right)^2}{y\left(y-1\right)\left(y+1\right)}\)
\(A=\frac{y-1}{y+1}\)
c)Đặt A=2\(\Rightarrow\frac{y-1}{y+1}=2\Rightarrow2\left(y+1\right)=y-1\Rightarrow y=-3\)