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Huong dẫn: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) ( n\(\in\)N*) áp dụng vào từng cái ngoặc
\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{20}\left(1+2+3+......+20\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+......+\frac{1}{20}.\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+......+\frac{21}{2}=\frac{2+3+4+.....+21}{2}=\frac{20.23}{2}=230\)
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(B=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(B=1-\frac{1}{2}.\left(1+2\right).2:2-\frac{1}{4}.\left(1+4\right).4:2-...-\frac{1}{20}.\left(1+20\right).20:2\)
\(B=1-3:2-5:2-...-21:2\)
\(B=1-3.\frac{1}{2}-5.\frac{1}{2}-...-21.\frac{1}{2}\)
\(B=1-\frac{1}{2}.\left(3+5+...+21\right)\)
Đặt C = 3 + 5 + ... + 21
Số số hạng của tổng C là: (21 - 3) : 2 + 1 = 10 (số)
=> C = (3 + 21) x 10 : 2 = 24 x 5 = 120
=> \(A=1-\frac{1}{2}.120\)
\(A=1-60=-59\)