Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\\ \Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\\ \Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\\ \Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\\ \Leftrightarrow x+2004=0\\ \Leftrightarrow x=-2004\)
a) \(\frac{1}{2010}\)và \(\frac{-7}{19}\)
Ta có : \(\frac{1}{2010}>0>\frac{-7}{19}\)
\(\Rightarrow\frac{1}{2010}>\frac{-7}{19}\)
b)\(\frac{497}{-499}\)và \(\frac{-2345}{2341}\)
Ta có : \(\frac{497}{-499}< -1< \frac{-2345}{2341}\)
\(\Rightarrow\frac{497}{-499}>\frac{-2345}{2341}\)
c)\(\frac{2000}{2001}\)và \(\frac{2001}{2002}\)
Ta có : \(\frac{2000}{2001}=1-\frac{1}{2001};\frac{2001}{2002}=1-\frac{1}{2002}\)
mà \(\frac{1}{2001}>\frac{1}{2002}\Rightarrow1-\frac{1}{2001}< 1-\frac{1}{2002}\)
\(\Rightarrow\frac{2000}{2001}< \frac{2001}{2002}\)
bài 1\(\dfrac{1}{2002}+\dfrac{2003\cdot2001}{2002}+2003=\dfrac{1+2003\cdot2001+2003\cdot2002}{2002}=\dfrac{1+2003\left(2001+2003\right)}{2002}=1+2003\cdot2=4007\)
câu3
a)VP=\(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)=VT
b)VP=VT\(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\)
So sánh\(A=\frac{2^{2006}+7}{2^{2004}+7}\)và\(B=\frac{2^{2003}+1}{2^{2001}+1}\)
A A > B
B A = B
C A < B
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!