Ta có: \(M=\frac{101^{102}+1}{101^{103}+1}\Rightarrow101M=\frac{101^{103}+101}{101^{103}+1}=1+\frac{100}{101^{103}+1}\)
\(N=\frac{101^{103}+1}{101^{104}+1}\Rightarrow101N=\frac{101^{104}+101}{101^{104}+1}=1+\frac{100}{101^{104}+1}\)
Vì \(1+\frac{100}{101^{103}+1}>1+\frac{100}{101^{104}+1}\) nên \(101M>101N\)
\(\Rightarrow M>N\)
Vậy M > N
$\frac{1}{101}$$+$$\frac{1}{102}$$+$$\frac{1}{103}$$+$ $.............$ $+$$\frac{1}{200}$
Chứng minh:
A= \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\)Chứng minh:
A=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+......+\frac{1}{109}+\frac{1}{200}>\frac{7}{12}\)
CÁC BẠN ƠI GIÚP MIK VỚI MAI MIK PHẢI NỘP RỒI
Chứng minh rằng:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Help me!!!!!!!
so sánh các phân số sau:
\(\frac{20^{100}-1}{20^{101}-1}\) và \(\frac{20^{101}-1}{20^{102}-1}\)
1+(-2)+(-3)+4+5+(-6)+(-7)+8+.....+99-100-101+102+103
$A$$=$$\frac{1}{101}$$+$$\frac{1}{102}$$+$$\frac{1}{103}$$+$ $.............$ $+$$\frac{1}{200}$
$CMR$: $a$, $A$$>$$\frac{7}{12}$
$b$, $A$$>$$\frac{5}{8}$
B=1/100^2+1/101^2+1/102^2+1/103^2+...+1/199^2. chứng minh 1/100<B<1/99
CMR: \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{5}{8}\)
