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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
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: \(\hept{\begin{cases}A=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\left(y\ne\pm1\right)\\B=\frac{y^2-y}{2y+1}=\frac{y\left(y-1\right)}{2y+1}\left(y\ne\frac{-1}{2}\right)\end{cases}}\)
\(\Rightarrow M=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\cdot\frac{y\left(y-1\right)}{2y+1}=\frac{\left(2y+1\right)\cdot y\cdot\left(y-1\right)}{\left(y-1\right)\left(y+1\right)\left(2y+1\right)}=\frac{y}{y+1}\)
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 :
\(\frac{2.\left(\frac{1}{2}\right)^2+\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}+\frac{2}{2}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{12}\)
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :
\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)