1.
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)
\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+....+\frac{102-100}{100.102}\)
\(=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{100}-\frac{1}{102}\right)\times\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{102}\right)\times\frac{1}{2}\)
\(=\frac{25}{51}\times\frac{1}{2}\)
\(=\frac{25}{102}\)
1,
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{100.102}\)
\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{102-100}{100.102}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\)
\(2A=\frac{1}{2}-\frac{1}{102}\)
\(2A=\frac{25}{51}\)
\(A=\frac{25}{102}\)
2,câu hỏi tương tự
\(a=\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{100.102}\)
\(2a=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{1}{100.102}\)
đặt a=1/2x4+1/4x6+...+1/100x102
a x2=(1/2x4+1/4x6+...+1/100x102)x2
a x2=2/2x4+2/4x6+...+2/100x102
a x2=1/2-1/4+1/4-1/6+...+1/100-1/102
khử đi ta còn
a x2=1/2-1/102
a x2=25/51
a=25/51 :2=25/102