Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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\(\Rightarrow\frac{y-1}{y+1}=2\Rightarrow2\left(y+1\right)=y-1\Rightarrow y=-3\)
a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)
b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)
Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)
c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)
Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)
\(a.\) Ta có: \(B=\frac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}=\frac{3y^3-\left(6y^2+y^2\right)+\left(2y+3y\right)-1}{2y^3+\left(3y^2-4y^2\right)-\left(6y-2y\right)+3}\)
\(B=\frac{3y^3-y^2-6y^2+2y+3y-1}{2y^2+3y^2-4y^2-6y+2y+3}=\frac{y^2\left(3y-1\right)-2y\left(3y-1\right)+\left(3y-1\right)}{y^2\left(2+3\right)-2y\left(2y+3\right)+\left(2y+3\right)}\)
\(B=\frac{\left(3y-1\right)\left(y-1\right)^2}{\left(2y+3\right)\left(y-1\right)^2}=\frac{3y-1}{2y+3}\)
\(b.\)Ta có: \(\frac{2B}{2y+3}=\frac{2.\frac{3y-1}{2y+3}}{2y+3}=\frac{\frac{2.\left(3y-1\right)}{2y+3}}{2y+3}=\frac{2.\left(3y-1\right)}{\left(2y+3\right)^2}\in Z\)
\(\Rightarrow\)\(2y+3\inƯ\left(2\right)\)mà \(Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
Vì \(2y+3\)là số nguyên lẻ \(\Rightarrow\)\(2y+3=-1\) hoặc \(2y+3=1\)
\(2y=\left(-1\right)-3=-4\) \(2y=1-3=-2\)
\(y=\left(-4\right)\div2=-2\) \(y=\left(-2\right)\div2=-1\)
Vậy để \(\frac{2B}{2y+3}\in Z\) thì \(y=-2\) hoặc \(y=-1\)
\(c.\)Để \(B\ge1\)\(\Rightarrow\)\(B-1\ge0\) hay \(\frac{3y-1}{2y+3}-1\ge0\)\(\Rightarrow\)\(\frac{y-4}{2y+3}\ge0\)
* Trường hợp 1: \(y-4\ge0\) và \(2y+3>0\)
\(\Rightarrow\) \(y\ge4\) \(\Rightarrow\) \(2y\)\(>-3\)
* \(\Rightarrow\)\(y\)\(>-\frac{3}{2}\)
Vậy \(y\ge4\)
* Trường hợp 2: \(y-4\)\(\le\)\(0\) và \(2y+3\) \(< 0\)
\(\Rightarrow\)\(y\le4\) \(\Rightarrow\)\(2y< 3\)
\(\Rightarrow\)\(y< \frac{3}{2}\)
Vậy \(y\le4\)