\(\frac{5}{7}\times\frac{10}{11}+\frac{5}{7}\times\frac{14}{11}-\frac{5}{7}\times\frac{17}{11}\)

\(=\frac{5}{7}\times\left(\frac{10}{11}+\frac{14}{11}-\frac{17}{11}\right)\)

\(=\frac{5}{7}\times\frac{7}{11}\)

\(=\frac{5}{11}\)

Ta có:

\(\frac{5}{7}\cdot\frac{10}{11}+\frac{5}{7}\cdot\frac{14}{11}-\frac{5}{7}\cdot\frac{17}{11}\)

\(=\frac{5}{7}\left(\frac{10}{11}+\frac{14}{11}-\frac{17}{11}\right)\)

\(=\frac{5}{7}\left(\frac{10+14-17}{11}\right)\)

\(=\frac{5}{7}\cdot\frac{7}{11}\)

\(=\frac{5}{11}\)

có cần giúp hum

\(-8x-4=12\Leftrightarrow-8x=16\Leftrightarrow x=-2\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{3}=\dfrac{3}{4}\left(x-\dfrac{5}{3}\right)+1\Leftrightarrow\dfrac{3}{4}\left(x-\dfrac{5}{3}\right)=1+\dfrac{4}{3}=\dfrac{7}{3}\)

\(\Leftrightarrow x-\dfrac{5}{3}=\dfrac{28}{9}\Leftrightarrow x=\dfrac{43}{9}\)

số xăng ô tô tiêu thụ trung bình :

580/40=14,5(l)

Kết quả = 14,5 l

a/b = 2022

a + b = 4046

a/b + 1 = 2023

a/b + b/b = 2023

a + b/b = 2023

=> 4046/b = 2023

b = 4046/2023

b = 2

=> a = 4046 - 2 = 4044

KL: b = 2; a = 4044

`Answer:`

\(B=\frac{101}{120}+\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)

\(=\frac{101}{120}+\frac{1}{12}+\frac{1}{36}+\frac{1}{72}+...+\frac{1}{2052}+\frac{1}{2280}\)

\(=\frac{101}{120}+\frac{1}{6}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{342}+\frac{1}{380}\right)\)

\(=\frac{101}{120}+\frac{1}{6}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(=\frac{101}{120}+\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=\frac{101}{120}+\frac{1}{6}.\left(1-\frac{1}{20}\right)\)

\(=\frac{101}{120}+\frac{1}{6}.\frac{19}{20}\)

\(=\frac{101}{120}+\frac{19}{120}\)

\(=1\)

`->` Chọn D.