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6 tháng 8 2018

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

mà x là số tựu nhiên => \(x\in\varnothing\)

29 tháng 3 2018

a) ta có \(\frac{-5}{6}\)\(\times\)\(\frac{120}{25}\)\(x\)<\(\frac{-7}{15}\)\(\times\)\(\frac{4}{9}\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2074074074\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2\)

mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)( -1;-2;-3)

b) ta có \(\left(\frac{-5}{3}\right)^3\)<\(x\)<\(\frac{-25}{35}\)\(\times\)\(\frac{-5}{6}\)\(\Rightarrow\)\(-4,62962963\)<\(x\)<\(0,5952380952\)

mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)(-4;-3;-2;-1;0)

ĐÚNG THÌ K CHO MK NHA

Mơn bạn nha 

forever young

>_<

!!!!!!!!!!

13 tháng 4 2020

2/7 + 7/-5 + -2/32 = -41/35

4/3 + | 5\-6 | + | -7/-6 | = 4/3 + 5/6 + 7/6 = 10/3

hok tốt

.............

13 tháng 4 2020

\(\frac{2}{7}+\frac{7}{-5}+\frac{-2}{35}\)

\(=\frac{2}{7}+\frac{-7}{5}+\frac{-2}{35}\)

\(=\frac{10}{35}+\frac{-49}{35}+\frac{-2}{35}\)

\(=\frac{-41}{35}\)

\(\frac{4}{3}+\left|\frac{5}{6}\right|+\left|\frac{-7}{-6}\right|\)

\(=\frac{4}{3}+\frac{5}{6}+\frac{7}{6}\)

\(=\frac{8}{6}+\frac{5}{6}+\frac{7}{6}\)

\(=\frac{20}{6}=\frac{10}{3}\)

2 tháng 4 2019

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

2 tháng 4 2019

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

23 tháng 3 2019

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

Ta có:

\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{25}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)

\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{101}>\frac{1}{6}+\frac{1}{25}-\frac{1}{100}=\frac{1}{6}+\frac{3}{100}>\frac{1}{6}\left(1\right)\)

\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)

Từ (1) và (2) suy ra:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(đpcm\right)\)

13 tháng 8 2019

đạt 1/52+.........+1/1002=S

1/52>1/5*6

.....................

1/1002>1/100*101

=>S>1/5*6+.............+1/100*101=1/5-1/6+....+1/100-1/101=1/5-1/101=96/505>96/576=1/6

 vậ S>1/6

1/52<1/4*5

.....................

1/1002<1/99*100

=>S<1/4*5+................+1/99*100=1/4-1/5+.....+1/99-1/100=1/4-1/100=6/25<6/24=1/4

 Vậy 1/6<S<1/4