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Bài 8:
a: \(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
b: \(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)\)
\(=2\cdot\dfrac{5}{90}=\dfrac{10}{90}=\dfrac{1}{9}\)
c: \(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)
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tạm thời cái đề bài là tìm x vậy
tiếp tục thui
=) (6x-10)-(6x-3)\(⋮\)2x-1
=)6x-10-6x+3\(⋮\)2x-1
=) (6x-6x)-(10-3)\(⋮\)2x-1
=)0-7\(⋮\)2x-1
=)-7\(⋮\)2x-1=)2x-1\(\in\)Ư(-7)={-7;-1;1;7}
=)2x\(\in\){-6;0;2;8}
=)x\(\in\){-3;0;1;4}
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(x-2)3=216
=>\(\left(x-2\right)^3=6^3\)
=>x-2=6
=>x=8
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b, (112005 + 112004) : 112003
= ( 112005 : 112003 ) : ( 112005 : 11 2003)
= (112005 - 2003) : ( 112004-2003)
=112 : 111 = 112-1=111=11
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`1,14 . 6,4 + 1,14 . 3,6 + 0,2 . 1,14 . 5`
`= 1,14 . (3,6 + 6,4 + 1)`
`= 1,14 . 11`
`= 12,54`
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ta có A=\(\dfrac{6}{8}\)+\(\dfrac{6}{56}\)+\(\dfrac{6}{140}\)+...+\(\dfrac{6}{1100}\)+\(\dfrac{6}{1400}\)
=\(\dfrac{3}{4}\)+\(\dfrac{3}{28}\)+\(\dfrac{3}{70}\)+...+\(\dfrac{3}{550}\)+\(\dfrac{3}{700}\)
=\(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{22.25}\)+\(\dfrac{3}{25.28}\)
=1-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+...+\(\dfrac{1}{22}\)-\(\dfrac{1}{25}\)+\(\dfrac{1}{25}\)-\(\dfrac{1}{28}\)
=1-\(\dfrac{1}{28}\)
=\(\dfrac{27}{28}\)
Vậy A=\(\dfrac{27}{28}\)
Ta có:
A =6/8+6/56+6/140+...+6/1100+6/1400
⇒A=3/4+3/28+3/70+...+3/550+3/700
⇒A=3/1.4+3/4.7+3/7.10+...+3/22.25+3/25.28
⇒A=1−1/4+1/4−1/7+1/7−1/10+...+1/22−1/25+1/25−1/28
⇒A=1−1/28
⇒A=1-1/38
\(\dfrac{4x}{3}=\dfrac{-8}{9}\)
\(4x=\dfrac{\left(-8\cdot3\right)}{9}\)
\(4x=-\dfrac{24}{9}\)
\(x=-\dfrac{24}{9}\div4\)
\(x=-\dfrac{24}{36}\)
\(x=-\dfrac{2}{3}\)