\(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)

= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{47}{30}\)

\(\dfrac{2}{6}+\dfrac{3}{12}\)

= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)

= \(\dfrac{7}{12}\)

\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)

= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)

= \(\dfrac{7}{6}\)

   \(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)

= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{19}{12}\)

    \(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)

= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)

=  \(\dfrac{67}{56}\) 

\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)

= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)

= \(\dfrac{73}{21}\)

   \(\dfrac{1}{5}+\dfrac{5}{35}\)

= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)

= \(\dfrac{12}{35}\) 

a: \(=\dfrac{-5}{8}+\dfrac{7}{12}=\dfrac{-15}{24}+\dfrac{14}{24}=\dfrac{-1}{24}\)

b: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}+\dfrac{11}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

c: \(=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\)

d: \(=\dfrac{5}{12}+\dfrac{5}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

e: \(=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\)

f: \(=\dfrac{-5}{15}+\dfrac{3}{25}=\dfrac{-25}{75}+\dfrac{9}{75}=\dfrac{-16}{75}\)

\(a,\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{12}{15}+\dfrac{10}{15}+\dfrac{1}{9}=\dfrac{22}{15}+\dfrac{1}{9}=\dfrac{66}{45}+\dfrac{5}{45}=\dfrac{71}{45}\)

\(b,\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}=\dfrac{53}{28}\)

\(c,\dfrac{1}{2}+\dfrac{1}{7}+\dfrac{-1}{5}=\dfrac{7}{14}+\dfrac{2}{14}+\dfrac{-1}{5}=\dfrac{9}{14}+\dfrac{-1}{5}=\dfrac{45}{70}+\dfrac{-14}{70}=\dfrac{31}{70}\)

\(d,\dfrac{7}{8}+\dfrac{5}{16}+\dfrac{-3}{4}=\dfrac{14}{16}+\dfrac{5}{16}+\dfrac{-12}{16}=\dfrac{7}{16}\)

\(e,\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{3}{12}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{8}{12}+\dfrac{-1}{13}=\dfrac{2}{3}+\dfrac{-1}{13}=\dfrac{26}{39}+\dfrac{-3}{39}=\dfrac{23}{39}\)

\(g,\dfrac{2}{3}+\dfrac{3}{8}+\dfrac{-5}{12}=\dfrac{16}{24}+\dfrac{9}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-10}{24}=\dfrac{15}{24}\)

de ma

c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

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

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

xcsxccsz

1 a)38000             b)1/8                c)3/2          d)9/17

2  a) 0.64           b)0.025                 c)-1

Bài giải:

Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)

\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)

\(=\left(-10\right).\left(-100\right).38\)

\(=1000.38=38000\)

b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)

\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)

\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)

c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)

\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)

\(=\frac{-5}{8}.1+\frac{17}{8}\)

\(=\frac{3}{2}\)

Câu 2: a, \(x-\frac{2}{5}=0,24\)

               \(x-0,4=0,24\)

               \(x=0,24+0,4\)

         \(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)

b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)

\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)

\(\frac{2}{3}.x=\frac{1}{60}\)

       \(x=\frac{1}{60}:\frac{2}{3}\)

   \(\Rightarrow x=\frac{1}{40}\)

c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)

\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)

\(\frac{7}{2}-2x=\frac{11}{2}\)

            \(2x=\frac{7}{2}-\frac{11}{2}\)

             \(2x=-2\)

            \(\Rightarrow x=-2:2\)

                  \(x=-1\)

thank

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

\(7,\)

\(A=7+\left(\dfrac{7}{12}-\dfrac{1}{2}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+\dfrac{7}{12}-\dfrac{1}{2}+3-\dfrac{1}{12}-5\)

\(=\left(\dfrac{7}{12}-\dfrac{1}{12}\right)+\left(7+3-5\right)-\dfrac{1}{2}\)

\(=\dfrac{6}{12}+5-\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+5\)

\(=5\)

\(8,\)

\(A=-\dfrac{1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{68}{49}\right)\)

\(=-\dfrac{1}{4}+\dfrac{7}{33}-\dfrac{5}{3}+\dfrac{15}{12}-\dfrac{6}{11}+\dfrac{68}{49}\)

\(=\left(-\dfrac{1}{4}+\dfrac{15}{12}\right)+\left(\dfrac{7}{33}-\dfrac{5}{3}-\dfrac{6}{11}\right)+\dfrac{68}{49}\)

\(=\left(-\dfrac{3}{12}+\dfrac{15}{12}\right)+\left(\dfrac{7}{33}-\dfrac{55}{33}-\dfrac{18}{33}\right)+\dfrac{68}{49}\)

\(=\dfrac{12}{12}-\dfrac{66}{33}+\dfrac{68}{49}\)

\(=1-2+\dfrac{68}{49}\)

\(=-1+\dfrac{68}{49}\)

\(=\dfrac{19}{49}\)

\(a,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)

\(=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{132}.\dfrac{132}{103}=\dfrac{115}{103}\)

\(b,\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2-5^0\)

\(=\dfrac{1}{4}-4-1=-\dfrac{19}{4}\)

\(c,12\dfrac{1}{3}-\dfrac{5}{7}:\left(24-23\dfrac{5}{7}\right)\)

\(=\dfrac{37}{3}-\dfrac{5}{7}.\dfrac{2}{7}=\dfrac{37}{3}-\dfrac{10}{49}=\dfrac{1783}{147}\)