a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{x}{9}=0\)

hay x=0

b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)

<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)

=> x-3=3

<=> x=6

Vậy x=6

\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)

* \(\dfrac{x}{15}=\dfrac{-2}{5}\)

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

\(\Rightarrow x=-6\)

*\(\dfrac{4}{y}=\dfrac{-2}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)

\(\Rightarrow y=-10\)

Vậy x = - 6 ; y = - 10

\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)

=> ( x - 3 ) . 20 = 4. 15

=> 20x - 60 = 60

=> 20x = 60 + 60

=> 20x = 120

=> x = 120 : 20

=> x = 6

Vậy x = 6

\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)

\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)

\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)

\(\Rightarrow-3+\left(-1\right)< x\le-1\)

\(\Rightarrow-4< x\le-1\)

\(\Rightarrow x=-3;-2;-1\)

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

\(\Leftrightarrow x=-\dfrac{3}{4}\)

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

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

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

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=-\dfrac{2}{3}\)

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

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

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

#YVA6

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=\dfrac{-2}{3}\)

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

\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)

\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

\(#yH\)

\(#NBaoNgoc\)

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

\(TH1:x\ge0\)

\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)

\(TH2:x< 0\)

\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)

Vậy ...

Giải:

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(TH1:x\ge0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

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

\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

         \(x.\dfrac{1}{6}=\dfrac{5}{6}\) 

              \(x=\dfrac{5}{6}:\dfrac{1}{6}\) 

              \(x=5\) 

\(TH2:x\le0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\) 

\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

     \(x.\dfrac{-11}{6}=\dfrac{5}{6}\) 

               \(x=\dfrac{5}{6}:\dfrac{-11}{6}\) 

               \(x=\dfrac{-5}{11}\) 

Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)

a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)

=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)

=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)

=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)

=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)

b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)

\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)

\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)

\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)

Bài 2:

a: -2*(-27)=54

6*9=54

=>Hai phân số này bằng nhau

b: -1/-5=1/5=5/25<>4/25

Bài 3:

a: =>16/x=-4/5

=>x=-20

b: =>(x+7)/15=-2/3

=>x+7=-10

=>x=-17

a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)

\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)

Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)

b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)

\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)

Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)

Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)

Bài 3

a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\left(x+7\right).36=15.\left(-24\right)\)

\(36x+252=-360\)

\(36x=-360-252\)

\(36x=-612\)

\(x=\dfrac{-612}{36}\)

\(x=-17\)