a, \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{13}{12}+\dfrac{13}{32}=\dfrac{143}{96}\)

b, \(\dfrac{5}{-8}+\dfrac{14}{39}-\dfrac{6}{10}\)

\(\dfrac{-5}{8}+\dfrac{14}{39}-\dfrac{3}{5}\)

\(=\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{14}{39}\)

\(=\dfrac{-49}{40}+\dfrac{14}{39}=\dfrac{-1351}{1560}\)

\(\dfrac{39}{x+2}=\dfrac{26}{12}\\ \Rightarrow26\times\left(x+2\right)=39\times12\\ \Rightarrow26\times\left(x+2\right)=468\\ \Rightarrow x+2=468:26\\ \Rightarrow x+2=18\\ \Rightarrow x=18-2\\ \Rightarrow x=16\)

Bài 1:

\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\) 

\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\) 

\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\) 

Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\) 

\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\) 

\(\dfrac{-18}{31}\) 

\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\) 

Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)

Bài 2:

a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\) 

b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)

a) \(\dfrac{20}{12}\)          b) \(\dfrac{15}{15}\)          c) \(\dfrac{24}{42}\)

d) \(\dfrac{26}{39}\)          e) \(\dfrac{0}{8}\)          g) \(\dfrac{15}{3}\)

a) \(\dfrac{15}{3}\)

b)\(\dfrac{15}{15}\)

c) \(\dfrac{24}{42}\)

d) \(\dfrac{26}{39}\)

e)\(\dfrac{0}{8}\)

f) \(\dfrac{15}{15}\)

Giải:

\(9-3\times\left(x-9\right)=6\) 

      \(3\times\left(x-9\right)=9-6\) 

      \(3\times\left(x-9\right)=3\) 

               \(x-9=3:3\) 

               \(x-9=1\) 

                     \(x=1+9\) 

                     \(x=10\) 

\(4+6\times\left(x+1\right)=70\) 

      \(6\times\left(x+1\right)=70-4\) 

      \(6\times\left(x+1\right)=66\) 

               \(x+1=66:6\) 

               \(x+1=11\) 

                     \(x=11-1\) 

                     \(x=10\) 

\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\) 

         \(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\) 

         \(\dfrac{x}{13}=\dfrac{4}{13}\) 

\(\Rightarrow x=4\) 

\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{3}{7}\) 

\(\Rightarrow x=7\) 

\(5\times\left(3+7\times x\right)=40\) 

         \(3+7\times x=40:5\) 

         \(3+7\times x=8\) 

                \(7\times x=8-3\) 

                \(7\times x=5\) 

                      \(x=5:7\) 

                      \(x=\dfrac{5}{7}\) 

\(x\times6+12:3=120\) 

       \(x\times6+4=120\) 

             \(x\times6=120-4\) 

             \(x\times6=116\) 

                   \(x=116:6\) 

                   \(x=\dfrac{58}{3}\) 

\(x\times3,7+x\times6,3=120\) 

    \(x\times\left(3,7+6,3\right)=120\) 

                  \(x\times10=120\) 

                           \(x=120:10\) 

                           \(x=12\) 

\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\) 

      \(\left(360-x\right):0,25=400\) 

                   \(360-x=400.0,25\) 

                   \(360-x=100\) 

                             \(x=360-100\) 

                             \(x=260\) 

\(71+65\times4=\dfrac{x+140}{x}+260\) 

\(\left(x+140\right):x+260=71+260\) 

\(x:x+140:x+260=331\) 

    \(1+140:x+260=331\) 

                    \(140:x=331-1-260\) 

                    \(140:x=70\) 

                             \(x=140:70\) 

                             \(x=2\) 

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\) 

                      \(10\times x+\left(1+4+7+...+28\right)=155\)

Số số hạng \(\left(1+4+7+...+28\right)\) :

         \(\left(28-1\right):3+1=10\) 

Tổng dãy \(\left(1+4+7+...+28\right)\) :

         \(\left(1+28\right).10:2=145\) 

\(\Rightarrow10\times x+145=155\) 

               \(10\times x=155-145\) 

               \(10\times x=10\) 

                       \(x=10:10\) 

                       \(x=1\) 

Đều theo cách lớp 5 nha em!

a: =(2/7-2/7)(-4/7-5/9)=0

b:

Sửa đề: 9/13*(-12/17)+9/13*29/27

=9/13(-12/17+29/17)

=9/13*17/17=9/13

c: \(=\dfrac{1}{7}\left(4+\dfrac{6}{7}+\dfrac{8}{7}\right)=\dfrac{1}{7}\cdot6=\dfrac{6}{7}\)

d: =7/10(5/7+9/7+8/7+13/7)

=5*7/10=7/2

a) \(\dfrac{-12}{15}+\dfrac{-4}{26}=\dfrac{-4}{5}+\dfrac{-2}{13}=\dfrac{-52-10}{65}=\dfrac{-62}{65}\)

b) \(5\dfrac{1}{3}-2\dfrac{4}{5}=\dfrac{16}{3}-\dfrac{14}{5}=\dfrac{80}{15}-\dfrac{42}{15}=\dfrac{38}{15}\)

c) \(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)+\dfrac{-5}{10}=\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{1}{2}=\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{35}{70}=\dfrac{41}{70}\)

d) \(-1\dfrac{2}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-9}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-54}{42}+\dfrac{9}{42}-\dfrac{10}{42}=\dfrac{-55}{42}\)

e) \(12-\dfrac{11}{121}+\left(\dfrac{-8}{9}\right)-\left(-\dfrac{3}{7}\right)\)

\(=12-\dfrac{11}{121}-\dfrac{8}{9}+\dfrac{3}{7}\)

\(=\dfrac{91476}{7623}-\dfrac{693}{7623}-\dfrac{6776}{7623}+\dfrac{3267}{7623}\)

\(=\dfrac{7934}{693}\)

Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)

Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)

Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)

Từ (1) và (2) ta suy ra ĐPCM