=>3/10(x-5)=2x+5
=>3/10x-3/2=2x+5
=>-17/10x=5+3/2=6/2+3/2=11/2
=>x=-55/17
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Giúp mình giải bài này vs:
\(\dfrac{30}{100}\).x+\(\dfrac{1}{4}\) = \(\dfrac{1}{5}\) .x-\(\dfrac{1}{2}\)
(\(\dfrac{1}{7.9}\) +\(\dfrac{1}{9.11}\) +........+\(\dfrac{1}{31.33}\)).x=(0,25-3,5).\(\dfrac{27}{3}\)
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CM\(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
Cho M = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\) ; N = \(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\).
Tính M \(\cdot\) N.
CM
\(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
\(\dfrac{3}{4!}+\dfrac{3}{5!}+...+\dfrac{3}{100!}< \dfrac{1}{3!}\)
cmr \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a)dfrac{1}{2};dfrac{1}{6};dfrac{1}{12};dfrac{1}{20};dfrac{1}{30};...
b)dfrac{1}{6};dfrac{1}{66};dfrac{1}{176};dfrac{1}{336};...
Bài 2: Tính:
a)Adfrac{1+dfrac{1}{3}+dfrac{1}{5}+...+dfrac{1}{97}+dfrac{1}{99}}{dfrac{1}{1.99}+dfrac{1}{3.97}+dfrac{1}{5.95}+...+dfrac{1}{97.3}+dfrac{1}{99.1}}
b)Bdfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{100}}{dfrac{99}{1}+dfrac{98}{2}+dfrac{97}{3}+...+dfrac{1}{99}}
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Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a)\(\dfrac{1}{2};\dfrac{1}{6};\dfrac{1}{12};\dfrac{1}{20};\dfrac{1}{30};...\)
b)\(\dfrac{1}{6};\dfrac{1}{66};\dfrac{1}{176};\dfrac{1}{336};...\)
Bài 2: Tính:
a)A=\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
b)B=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}}\)
20 tháng 3 2021 lúc 16:09
Giúp mk vớiCâu 1:Cho A dfrac{1}{dfrac{99}{dfrac{1}{2}+}}+dfrac{2}{dfrac{98}{dfrac{1}{3}+}}+dfrac{3}{dfrac{97}{dfrac{1}{4}+....}}+...+dfrac{99}{dfrac{1}{dfrac{1}{100}}}. B dfrac{92}{dfrac{1}{45}+}-dfrac{1}{dfrac{9}{dfrac{1}{50}+}}-dfrac{2}{dfrac{10}{dfrac{1}{55}+}}-dfrac{3}{dfrac{11}{dfrac{1}{60}+....}}-...dfrac{92}{dfrac{100}{dfrac{1}{500}}}. Tính dfrac{A}{B}
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Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
10 tháng 4 2021 lúc 22:28
Tìm x∈Z, biết:a) left(x-20right)+left(x-19right)+left(x-18right)+...+99+100100b) 213-x.left(dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{2020}}right):left(1-dfrac{1}{2^{2020}}right)13
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\(Tìm\) \(x\)∈\(Z\)\(,\) \(biết\)\(:\)
\(a\)) \(\left(x-20\right)+\left(x-19\right)+\left(x-18\right)+...+99+100=100\)
\(b\)) \(213-x.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right):\left(1-\dfrac{1}{2^{2020}}\right)=13\)