4/3x6 + 4/6x9 + 4/9x12 +4/12x15
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3/3x6 + 3/6x9 + 3/9x12 + 3/12x15 + 3/15x18
= 1/3 - 1/6 + 1/6 - 1/9 + ... + 1/15 - 1/18
= 1/3 - 1/18
= 5/18
\(\frac{3}{3\times6}+\frac{3}{6\times9}+\frac{3}{9\times12}+\frac{3}{12\times15}+\frac{3}{15\times18}\)
=\(3\times\left(\frac{1}{3\times6}+\frac{1}{6\times9}+\frac{1}{9\times12}+\frac{1}{12\times15}+\frac{1}{15\times18}\right)\)
=\(3\times\left(\frac{1}{3}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{9}+\frac{1}{9}\times\frac{1}{12}+\frac{1}{12}\times\frac{1}{15}+\frac{1}{15}\times\frac{1}{18}\right)\)
=\(3\times\left(\frac{1}{3}\times\frac{1}{18}\right)\)
=\(3\times\frac{1}{54}=\frac{3}{54}=\frac{1}{18}\)
A = \(\dfrac{7}{3\times6}\) + \(\dfrac{7}{6\times9}\) + \(\dfrac{7}{9\times12}\) + \(\dfrac{7}{12\times15}\)+ .....+\(\dfrac{7}{96\times99}\)
A = \(\dfrac{7}{3}\) x ( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+ \(\dfrac{3}{9\times12}\)+ \(\dfrac{3}{12\times15}\)+......+\(\dfrac{3}{96\times99}\))
A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{15}\)+....+ \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\))
A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\)- \(\dfrac{1}{99}\))
A = \(\dfrac{224}{297}\)
Ta có: \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^6}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^6}}{2}\)
A = \(\dfrac{4}{3\times6}\) + \(\dfrac{4}{6\times9}\) + ......+ \(\dfrac{4}{18\times21}\)
A = \(\dfrac{4}{3}\) \(\times\) ( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+......+ \(\dfrac{3}{18\times21}\))
A = \(\dfrac{4}{3}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{9}\)+.....\(\dfrac{1}{18}\) - \(\dfrac{1}{21}\))
A = \(\dfrac{4}{3}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{21}\))
A = \(\dfrac{4}{3}\)\(\times\) \(\dfrac{2}{7}\)
A = \(\dfrac{8}{21}\)
nhân cả vế với 3 ta có
Ax3=\(\frac{3}{3x6}\)+\(\frac{3}{6x9}\)+.........+\(\frac{3}{99x102}\)
Ax3=\(\frac{1}{3}\)-\(\frac{1}{6}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{102}\)
Ax=\(\frac{1}{3}\)-\(\frac{1}{102}\)
Ax3=\(\frac{11}{34}\)
A=\(\frac{11}{34}\):3
A=\(\frac{11}{102}\)
gạch đi các số lặp lại thì còn phân số 1/3 và 1/102 lấy \(\frac{1}{3}-\frac{1}{102}=\frac{33}{102}\)
Chọn C
6 x 9 - 2 x 6 + 8 x 3 : 2 x 3
= 6 x 9 : 2 x 3 + - 2 x 6 : 2 x 3 + 8 x 3 : 2 x 3
= 3 x 6 - x 3 + 4
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)
\(A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{93.96}+\frac{3}{96.99}\)
\(A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{93}-\frac{1}{96}+\frac{1}{96}-\frac{1}{99}\)
\(A=1-\frac{1}{99}=\frac{98}{99}\)
Vậy A=\(\frac{98}{99}\)
\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(3B=\)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(3B=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)
\(B=\frac{24}{49}:3=\frac{8}{49}\)
Vậy B=\(\frac{8}{49}\)
Dấu "." là dấu nhân.
_Học tốt_
= 73/3 nhé