Ta có \(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+...\frac{1}{2002}=VP\)
Vậy...
Tìm giá trị nguyên của x và y thỏa mãn: 3xy+x-y=1
CMR: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
chứng minh : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+.....+\frac{1}{2002}\)
Biết
1)\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\).CM: \(a^2=b\cdot c\)
2)CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
A = \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+\(\frac{1}{2001^2}\)+\(\frac{1}{2002^2}\)
Chứng minh rằng A<\(\frac{2001}{2002}\)
\(\frac{y-1}{2004}+\frac{y-2}{2003}-\frac{y-3}{2002}=\frac{y-4}{2001}\)
\(x+\frac{4}{2000}+x+\frac{3}{2001}=x+\frac{2}{2002}+x+\frac{1}{2003}\)
\(x+\frac{4}{2000}+x+\frac{3}{2001}=x+\frac{2}{2002}+x+\frac{1}{2003}\)
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{2001^2}\)+\(\frac{1}{2002^2}\)
thực hiện phép tính:
a)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b)\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
