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18 tháng 7 2018

\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)

\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\) 

AH
Akai Haruma
Giáo viên
16 tháng 7 2018

Lời giải:

\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)

\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)

\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)

26 tháng 3 2017

Ta gọi biểu thức đó là D

\(D=\frac{5}{2}\left[\frac{1}{5.8}-\frac{1}{8.11}+....+\frac{1}{302.305}-\frac{1}{305.308}\right]\)

\(D=\frac{5}{2}.\left[\frac{1}{5.8}-\frac{1}{305.308}\right]\)

\(D=\frac{4695}{75152}\)

21 tháng 8 2017

a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0

22 tháng 8 2017

a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran

b: \(x+1⋮x-2\)

\(\Leftrightarrow x-2+3⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

c: \(\Leftrightarrow2x-1\in\left\{12;24;36;48\right\}\)

\(\Leftrightarrow x\in\left\{\dfrac{13}{2};\dfrac{25}{2};\dfrac{37}{2};\dfrac{49}{2}\right\}\)

d: \(\Leftrightarrow n+3+2⋮n+3\)

\(\Leftrightarrow n+3\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{-2;-4;-1;-5\right\}\)

e: \(2n-1⋮3n+6\)

\(\Leftrightarrow6n-3⋮3n+6\)

\(\Leftrightarrow6n+12-15⋮3n+6\)

\(\Leftrightarrow3n+6\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(n\in\left\{-\dfrac{5}{3};-\dfrac{7}{3};-1;-3;-\dfrac{1}{3};-\dfrac{11}{3};3;-7\right\}\)