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\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=1\cdot3\cdot2\)
\(=6\)
\(A^2=6^2=36\)
Ta có: \(\frac{1}{4\times6}=\frac{1}{4\times1\times3\times2}=\frac{1}{4\times3\times1\times2}\)
\(\frac{1}{8\times9}=\frac{1}{4\times2\times3\times3}=\frac{1}{4\times3\times2\times3}\)
\(\frac{1}{12\times12}=\frac{1}{4\times3\times3\times4}\)
\(\frac{1}{16\times15}=\frac{1}{4\times4\times3\times5}=\frac{1}{4\times3\times4\times5}\)......
\(\frac{1}{2680\times2013}=\frac{1}{4\times670\times3\times671}\)
Do đó:
\(M=\frac{1}{4\times3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{670\times671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{670}-\frac{1}{671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{671}\right)=\frac{1}{12}\times\frac{670}{671}=\frac{335}{4026}\)
Vậy \(M=\frac{335}{4026}\)
4/3 x (1/5 + 1/4) + 4/3 x (3/16 - 3/4)
= 4/3 x (1/5 + 1/4 + 3/16 - 3/4)
= 4/3 x [1/5 + 3/16 - (3/4 - 1/4)]
= 4/3 x [1/5 + 3/16 - 1/2]
= 4/3 x [16/80 + 15/80 - 40/80]
= 4/3 x [- 9/80]
= - 3/20
Câu 2:
\(\frac{3^7.5^5}{9^2.5^2}\)
= \(\frac{3^7.5^5}{\left(3^2\right)^2.5^2}\)
= \(\frac{3^7.5^5}{3^4.5^2}\)
= 3^3.5^3
= 15^3
= 3375
nhầm = 16/45
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
giải:
\(\frac{4}{3x6}\)+\(\frac{4}{6x9}\)+\(\frac{4}{9x12}\)+ \(\frac{4}{12x15}\)
= \(\frac{4}{3}\)x(\(\frac{3}{3x6}\)+ \(\frac{3}{6x9}\)+\(\frac{3}{9x12}\)+\(\frac{3}{12x15}\))
=\(\frac{4}{3}\)x(1-\(\frac{1}{15}\))
=\(\frac{4}{3}\)x\(\frac{14}{15}\)
=\(\frac{56}{45}\)