a) 4/9 x ( 3/5 + 8/5 - 2/10 )

= 4/9 x 1

=4/9

b) ( 312 + 325 - 247 ) : 13

= 390 : 13

= 30

a)4/9 x 3/5 + 4/9 x 8/5 - 4/9 x 2/10 

=4/9x(3/5+8/5-2/10)

= 4/9x1

=4/9

b,312 : 13 + 325 : 13 - 247 : 13

=(312+325-247) : 13

=         390 : 3

= 130

a: \(=\dfrac{7+12-6}{13}=1\)

b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)

c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)

d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)

Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.

a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.

b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.

c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao? 
cho mìn hỏi câu b nhoa

a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\) 

\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)

\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)

b: =>2(x+7/8)=6*13/4=78/4=39/2

=>x+7/8=39/4

=>x=71/8

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{17}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{13}\)

\(\dfrac{-1}{4}+\dfrac{3}{4}-x=\dfrac{-5}{4}\)

\(< =>\dfrac{2}{4}-x=\dfrac{-5}{4}\)

\(< =>x=\dfrac{2}{4}-\dfrac{-5}{4}=\dfrac{7}{4}\)

\(\dfrac{2}{13}-x=\dfrac{-5}{13}\)

\(< =>x=\dfrac{2}{13}-\dfrac{-5}{13}=\dfrac{7}{13}\)

\(\dfrac{-1}{4}+\dfrac{3}{4}-x=\dfrac{-5}{4}\)

\(\Leftrightarrow\dfrac{1}{2}-x=\dfrac{-5}{4}\)

\(\Rightarrow x=\dfrac{7}{4}\)

\(\dfrac{-1}{4}+\dfrac{3}{4}-x=\dfrac{-5}{4}\)

           \(\dfrac{2}{4}-x=\dfrac{-5}{4}\)

                  \(x=\dfrac{-5}{4}-\dfrac{2}{4}\)

                  \(x=-\dfrac{7}{4}\)

\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

\(\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{60}{140}+\dfrac{56}{140}+\dfrac{105}{140}=\dfrac{221}{140}\)

\(\dfrac{9}{7}-\dfrac{5}{11}x\dfrac{11}{7}=\dfrac{9}{7}-\dfrac{5}{7}=\dfrac{4}{7}\)

a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)

b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)

c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)

d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)

\(a,\)\(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(x=\dfrac{13}{7}.\dfrac{6}{13}\)

\(x=\dfrac{6}{7}\)

b,\(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)

\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)

\(\dfrac{4}{7}.x=\dfrac{13}{15}\)

\(x=\dfrac{13}{15}:\dfrac{4}{7}\)

\(x=\dfrac{13}{15}.\dfrac{7}{4}\)

\(x=\dfrac{91}{60}\)

a: Ta có: \(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(\Leftrightarrow x=\dfrac{13}{7}\cdot\dfrac{6}{13}\)

hay \(x=\dfrac{6}{7}\)

b: Ta có: \(\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{13}{15}\)

hay \(x=\dfrac{91}{60}\)

\(=\dfrac{29}{13}\cdot\dfrac{13}{29}\cdot4\cdot\dfrac{21}{8}\cdot\dfrac{8}{21}=4\)

bài này mik ra \(\dfrac{39}{29}\) ko bt đúng ko :)