Câu 1 :
\(x:\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{101.103}\right)=1\)
\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\right]\) \(=1\)
\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{103}\right)\right]=1\)
\(=>\) \(x:\frac{51}{103}=1\)
\(=>x=1.\frac{51}{103}=\frac{51}{103}\)
Câu 2 :
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\right).x=2\)
\(=>\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right).x=2\)
\(=>\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)
\(=>\frac{11}{12}.x=2\)
\(=>x=2:\frac{11}{12}\)
\(=>x=\frac{24}{11}\)
cau b bang 24/11
Cau a bang 51/103