40-5.6-10=?
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\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1\)
\(x=29\)
Vậy ...
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\(=\frac{2^{30}.3^{18}.2.3.7-3.2^{30}.3^{20}}{2^8.5^2.2^{20}.3^{20}-2^{10}.2^{19}.3^{19}.5.7}\)
\(=\frac{2^{31}.3^{19}.7-2^{30}.3^{21}}{2^{28}.5^2.3^{20}-2^{29}.3^{19}.5.7}\)
\(=\frac{2^{30}.3^{19}\left(2.7-3^2\right)}{2^{28}.3^{19}\left(5^2.3-2.5.7\right)}\)
\(=\frac{2^2.5}{5}=\frac{20}{5}=4\)
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\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)
n + 1 = 30
n = 30 - 1
n = 29
Kết luận n = 29 là giá trị thỏa mãn yêu cầu đề bài.
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\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{-1}{\left(n+1\right)}=\dfrac{-1}{30}\)
\(-n-1=-30\)
-n = -29
n = 29
40 - 5 x 6 - 10 = 40 - 30 - 10 = 40 - (30 + 10) = 40 - 40 = 0
=200
jk mik nhé