1/4 mũ 2+1/5 mũ 2 +1/6 mũ 2+...+1/100 mũ 2 < 1/3
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt số chia là a và thương là b ta có
\(\dfrac{200-13}{a}=b\Rightarrow b=\dfrac{187}{a}\Rightarrow187⋮a\)
\(\Rightarrow a=\left\{1;11;17\right\}\Rightarrow b=\left\{187;17;11\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Hiệu vận tốc của 2 xe là:
12-11=1 (km/giờ)
Hai xe gặp nhau sau thời gian là:
7:1=7 (giờ)
Hai xe gặp nhau lúc:
7 giờ + 7 giờ = 14 giờ
Chỗ gặp nhau cách A là:
12.7=84 (km)
Đáp số: 2 xe gặp nhau lúc 14 giờ và chỗ gặp nhau cách A 84 km
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{13}\cdot\dfrac{-3}{10}\cdot\dfrac{-3}{5}\)
\(=\dfrac{5}{5}\cdot\dfrac{-3}{13}\cdot\dfrac{-3}{10}\)
\(=\dfrac{3}{13}\cdot\dfrac{3}{10}=\dfrac{9}{130}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{4}\cdot\dfrac{1}{13}-\dfrac{3}{4}\cdot\dfrac{14}{13}\)
\(=\dfrac{3}{4}\left(\dfrac{1}{13}-\dfrac{14}{13}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{-13}{13}=-\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{10}{-13}:\dfrac{-4}{13}\cdot\dfrac{11}{-10}\)
\(=\dfrac{10}{13}\cdot\dfrac{13}{4}\cdot\dfrac{11}{-10}\)
\(=\dfrac{10}{4}\cdot\dfrac{11}{-10}=-\dfrac{11}{4}\)
\(\dfrac{10}{-13}:\dfrac{-4}{13}.\dfrac{11}{-10}\)
\(=\dfrac{10}{-13}.\dfrac{13}{4}.\dfrac{11}{-10}\)
\(=\dfrac{1430}{520}\)
\(=\dfrac{11}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{15}=\dfrac{-6}{5}\)
\(\Rightarrow x=\dfrac{-6}{5}\cdot15\)
\(\Rightarrow x=-6\cdot3\)
\(\Rightarrow x=-18\)
\(\dfrac{x}{15}=\dfrac{-6}{5}\\ \dfrac{x}{15}=\dfrac{-18}{15}\)
⇒\(x=-18\)
Vậy \(x=-18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{-2}{5}< \dfrac{x}{15}< \dfrac{1}{6}\)
\(\Rightarrow\dfrac{-2\cdot6}{5\cdot6}< \dfrac{2\cdot x}{15\cdot2}< \dfrac{5\cdot1}{6\cdot5}\)
\(\Rightarrow-12< 2x< 5\)
\(\Rightarrow-6< x< \dfrac{5}{2}\)
Mà: x nguyên
\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)
b) \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{12}< 2x< \dfrac{-12}{31}+\dfrac{-136}{-31}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{5}{12}< 2x< \dfrac{-12}{31}+\dfrac{136}{31}\)
\(\Rightarrow\dfrac{2}{12}-\dfrac{5}{12}< 2x< \dfrac{124}{31}\)
\(\Rightarrow-\dfrac{3}{12}< 2x< 4\)
\(\Rightarrow-\dfrac{1}{4}< x< 2\)
Mà x nguyên
\(\Rightarrow x\in\left\{0;1\right\}\)
a)\(\dfrac{-2}{5}< \dfrac{x}{15}< \dfrac{1}{6}\)
⇒\(\dfrac{-12}{30}< \dfrac{2.x}{30}< \dfrac{5}{30}\)
⇒\(-12< 2.x< 5\)
⇒\(2.x\)ϵ{\(-10;-8;-6;-4;-2;0;2;4\)}
⇒\(x\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\)
\(=\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...++\dfrac{100-99}{99.100}=\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)