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

\(\dfrac{5}{13}\cdot\dfrac{-3}{10}\cdot\dfrac{-3}{5}\)

\(=\dfrac{5}{5}\cdot\dfrac{-3}{13}\cdot\dfrac{-3}{10}\)

\(=\dfrac{3}{13}\cdot\dfrac{3}{10}=\dfrac{9}{130}\)

5/3 . (-3/10).(-3/5) 

=5/3 . 3/10 . 3/5

=45/150

=3/10

\(\dfrac{3}{4}\cdot\dfrac{1}{13}-\dfrac{3}{4}\cdot\dfrac{14}{13}\)

\(=\dfrac{3}{4}\left(\dfrac{1}{13}-\dfrac{14}{13}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{-13}{13}=-\dfrac{3}{4}\)

3/4×1/13-3/4×14/13

=3/4×(1/13-14/13)

=3/4×(-13/13)

=3/4×(-1)

=(-3/4)

\(\dfrac{10}{-13}:\dfrac{-4}{13}\cdot\dfrac{11}{-10}\)

\(=\dfrac{10}{13}\cdot\dfrac{13}{4}\cdot\dfrac{11}{-10}\)

\(=\dfrac{10}{4}\cdot\dfrac{11}{-10}=-\dfrac{11}{4}\)

\(\dfrac{10}{-13}:\dfrac{-4}{13}.\dfrac{11}{-10}\)

\(=\dfrac{10}{-13}.\dfrac{13}{4}.\dfrac{11}{-10}\)

\(=\dfrac{1430}{520}\)

\(=\dfrac{11}{4}\)

\(\dfrac{x}{15}=\dfrac{-6}{5}\)

\(\Rightarrow x=\dfrac{-6}{5}\cdot15\)

\(\Rightarrow x=-6\cdot3\)

\(\Rightarrow x=-18\)

\(\dfrac{x}{15}=\dfrac{-6}{5}\\ \dfrac{x}{15}=\dfrac{-18}{15}\)

⇒\(x=-18\)

Vậy \(x=-18\)

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{\(-10;-8;-6;-4;-2;0;2;4\)}

⇒\(x\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)

\(2-x=\dfrac{6}{5}\)

\(x=2-\dfrac{6}{5}\)

\(x=\dfrac{10}{5}-\dfrac{6}{5}\)

\(x=\dfrac{10-6}{5}\)

\(x=\dfrac{4}{5}\)

\(2-x=\dfrac{6}{5}\)

      \(x=2-\dfrac{6}{5}\)

      \(x=\dfrac{4}{5}\)

Vậy \(x=\dfrac{4}{5}\)

Phân số chỉ 35 mét đường còn lại là:

\(1-\dfrac{5}{9}-\dfrac{1}{4}=\dfrac{7}{36}\)

Đoạn đường cần sửa chữa dài:
\(35:\dfrac{7}{36}=180\left(m\right)\)

Đáp số: 180 m

ai trả lời giúp mik ik ạ !

Ta có:
\(\dfrac{2022}{2021}=\dfrac{2021+1}{2021}=1+\dfrac{1}{2021}\)

\(\dfrac{2021}{2020}=\dfrac{2020+1}{2020}=1+\dfrac{1}{2020}\)

Mà: \(2021>2020\)

\(\Rightarrow\dfrac{1}{2021}< \dfrac{1}{2020}\)

\(\Rightarrow1+\dfrac{1}{2021}< 1+\dfrac{1}{2020}\)

\(\Rightarrow\dfrac{2022}{2021}< \dfrac{2021}{2020}\)

\(\dfrac{2022}{2021}=1+\dfrac{1}{2021}\)

\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)

Do \(2021>2020\Rightarrow\dfrac{1}{2021}< \dfrac{1}{2020}\)

\(\Rightarrow1+\dfrac{1}{2021}< 1+\dfrac{1}{2020}\)

Vậy \(\dfrac{2022}{2021}< \dfrac{2021}{2020}\)