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\(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+...+\frac{7}{17\cdot21}=\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{17\cdot21}=\)
\(\frac{4}{7}A=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{17}-\frac{1}{21}\right)=\)
\(\frac{4}{7}A=1-\frac{1}{21}=\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}\div\frac{4}{7}\)
\(A=\frac{20}{21}\times\frac{7}{4}=\frac{140}{84}=\frac{5}{3}\)
\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+...+\frac{7}{17.21}\)
\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\frac{20}{21}=\frac{5}{3}\)
5 x 5 x 55 + 5 + 999999
= 25 x 55 + 5 + 999999
= 1375 + 5 + 999999
= 1380 + 999999
= 1001379
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)
Lời giải:
Gọi biểu thức số 1 là A và số 2 là B
\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)
5x5=25
4x5=20
1x5=5
5x5=25
4x5=20
1x5=5
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