1 + 1 + 2 =
Những câu hỏi liên quan
A=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+...(1000) số 1
B=2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+...+2(13512 số 2)
tổng của A và B là
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Tính tổng sau: a) 1/2+1/6+1/12+1/20+1/30 b) 1/15+1/35+1/63+1/99+1/143 c) 1/6+1/12+1/20+1/30+1/42+1/56 d) 1/2+1/2^2+1/2^3+1/2^4+1/2^5 e) 1/7+1/7^2+1/7^3+...+1/7^100 f) 1+1/2*(1+2)+1/3*(1+2+3)+1/4*(1+2+3+4)+...+1/200*(1+2+3+..+200) g) (1/2+1)*(1/3+1)*(1/4+1)*..*(1/100+1) h) (1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/2022) Giúp mk vs ạkkk
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
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1+1+1+1+1+1+1+1+1+1+1+1+11++1+1+1+11+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-2+2+2+2+2+2+2-12221+21
1+1+1+1+1+1+1+1+1+1+1+1+11++1+1+1+11+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-2+2+2+2+2+2+2-12221+21=
-12135
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Lớp một ai học cái này
Chứng Minh Rằng:
P=(1+1/a^2+1/(a+1)^2)^1/2=1+1/a-1/a-1
Tính :S=(1+1/1^2+1/2^2)^1/2+(1+1/2^2+3^2)^1/2+...+(1+99^2+100^2)^1/2
1+1+1+1+1+1+1+1+1+1+1:0=?
2+2+2+2+2+2+2+2+2+2×2÷1=?
a) không tính đươc vì ko có kết quả của 1:0
b) 22
nhớ tkkkkkkkkkkkk
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1+0+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+11+1+1++1+11+1+1+1+1+1+1+1+2+2+2+2+2+2222+33333+444+555+666+777+888+999-{1+0+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+11+1+1++1+11+1+1+1+1+1+1+1+2+2+2+2+2+2222+33333+444+555+666+777+888+999-}
lớp 1 học cái này à:(((??
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G=(1-1/2^2).(1-1/3^2).(1-1/4^2).......(1-1/100^2)
H=(1-1/2^2).(1-1/3^2).(1-1/4^2)........(1-1/2018^2)