Tính giá trị các biểu thức :
\(A=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\) \(B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
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\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
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\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
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`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
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1.tính giá trị biểu thức
a) dfrac{2^2}{1.3}+dfrac{3^2}{2.4}+dfrac{4^2}{3.5}+dfrac{5^2}{4.6}+dfrac{6^2}{5.7}
b) left(1+dfrac{1}{1.3}right).left(1+dfrac{1}{2.4}right).left(1+dfrac{1}{3.5}right).left(1+dfrac{1}{9.11}right)
2. Chứng tỏ:
dfrac{1}{201}+dfrac{1}{202}+.........+dfrac{1}{399}+dfrac{1}{400}dfrac{1}{2}
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1.tính giá trị biểu thức
a) \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
b) \(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).\left(1+\dfrac{1}{9.11}\right)\)
2. Chứng tỏ:
\(\dfrac{1}{201}+\dfrac{1}{202}+.........+\dfrac{1}{399}+\dfrac{1}{400}\)>\(\dfrac{1}{2}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
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Chứng tỏ :a, A dfrac{1}{2.4}+dfrac{1}{4.6}+...+dfrac{1}{2022.2024} dfrac{1}{4}b, B dfrac{1}{1.3}+dfrac{1}{3.5}+...+dfrac{1}{2013.2015} dfrac{1}{2}c, C dfrac{1}{3^2}+dfrac{1}{5^2}+dfrac{1}{7^2}+...+dfrac{1}{2013^2} dfrac{1}{4}d, D dfrac{1}{2^2}+dfrac{1}{4^2}+dfrac{1}{6^2}+...+dfrac{1}{2014^2} dfrac{1}{2}
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Chứng tỏ :
a, A = \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2022.2024}\) < \(\dfrac{1}{4}\)
b, B =\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}< \dfrac{1}{2}\)
c, C =\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{4}\)
d, D =\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{2}\)
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
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15 tháng 4 2023 lúc 13:47
a) dfrac{2}{1.2.3}+dfrac{2}{2.3.4}+dfrac{2}{3.4.5}+...+dfrac{2}{18.19.20}b) dfrac{4}{1.3.5}+dfrac{4}{3.5.7}+dfrac{4}{5.7.9}+...+dfrac{4}{21.23.25}c) dfrac{3}{1.2}-dfrac{5}{2.3}+dfrac{7}{3.4}-dfrac{9}{4.5}+...+dfrac{39}{19.20}-dfrac{41}{20.21}d) dfrac{8}{9}cdotdfrac{15}{16}cdotdfrac{24}{25}cdot...cdotdfrac{99}{100}cdotdfrac{120}{121}e) left(1+dfrac{7}{9}right)left(1+dfrac{7}{20}right)left(1+dfrac{7}{33}right)left(1+dfrac{7}{48}right)...left(1+dfrac{7}{180}right)Các bạn không nhất thiết phải làm h...
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a) \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
b) \(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+...+\dfrac{4}{21.23.25}\)
c) \(\dfrac{3}{1.2}-\dfrac{5}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+...+\dfrac{39}{19.20}-\dfrac{41}{20.21}\)
d) \(\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{120}{121}\)
e) \(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
Các bạn không nhất thiết phải làm hết, làm cho nó dễ hiểu được thì càng tốt để mk vận dụng
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
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Tính:A=\(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{2016.2018}{2017^2}\)
1.Tính
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
B=\(\dfrac{3}{5.6}+\dfrac{3}{6.7}+\dfrac{3}{7.8}+.....+\dfrac{3}{101.102}\)
C=\(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}\)
D=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
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1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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Bài 1 : Tìm 2 số biết hiệu của chúng bằng 5 và 50% số lớn = 1 nửa số bé.
Bài 2 : tính giá trị biểu thức: A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
Bài 3 : Tìm x
a , \(\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}\)
bài 2:
\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
bài 3:
\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)
\(=>x=3\)
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Thực hiện phép tính
a) A= \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)\)\(+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)
b) B=\(\dfrac{1-3}{1.3}+\dfrac{2-4}{2.4}+\dfrac{3-5}{3.5}+\dfrac{4-6}{4.6}+...+\dfrac{2011-2013}{2011.2013}+\dfrac{2012-2014}{2012.2014}-\dfrac{2013+2014}{2013.2014}\)
17 tháng 3 2023 lúc 20:16
Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
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