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Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
a,
= 44.(82+18)-4oo
= 44.100-400
= 4400-400
= 4000
b,
= [319+(-219)]+[598+(-98)
=100+500
=600
c,
= (17/28+18/29-19/30-20/31).0
=0
k cho mik nhé
\(M=1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
\(\Leftrightarrow M=1+\frac{1}{3-3^2+...+3^{19}-3^{20}}\)
Đặt A = 3 - 32 + ....+ 319 - 320
=> \(3A=3^2-3^3+...+3^{20}-3^{21}\)
\(\Rightarrow3A+A=3-3^{21}\)
\(\Rightarrow4A=3-3^{21}\)
\(\Rightarrow A=\frac{3-3^{21}}{4}\)
\(\Rightarrow M=1+\frac{1}{\frac{3-3^{21}}{4}}\)
!!!! K chắc lm linh tinh thôi
Sai thì sr nha
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)
b) \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
=> \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}\)
c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\frac{4}{21}=11\)
d.e) ktra lại đề
b = 20 thì ta thay 20 vào vị trí của b trong biểu thức.
(–1) . (–2) . (–3) . (–4) . (–5) . b
= (–1) . (–2) . (–3) . (–4) . (–5) . 20
= – (1 . 2 . 3 . 4 . 5 . 20)
= – 2400
ta có:1/n(1+2+...+n)=1/n.n((n+1))/2=(n+1)/2
=>S=1+3/2+2+5/2+...+10=43