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\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow94< x< 92\)
mà x là số tựu nhiên => \(x\in\varnothing\)
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
2/7 + 7/-5 + -2/32 = -41/35
4/3 + | 5\-6 | + | -7/-6 | = 4/3 + 5/6 + 7/6 = 10/3
hok tốt
.............
\(\frac{2}{7}+\frac{7}{-5}+\frac{-2}{35}\)
\(=\frac{2}{7}+\frac{-7}{5}+\frac{-2}{35}\)
\(=\frac{10}{35}+\frac{-49}{35}+\frac{-2}{35}\)
\(=\frac{-41}{35}\)
\(\frac{4}{3}+\left|\frac{5}{6}\right|+\left|\frac{-7}{-6}\right|\)
\(=\frac{4}{3}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{8}{6}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{20}{6}=\frac{10}{3}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
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:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(đpcm\right)\)
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