\(x+7\frac{4}{5}=9\frac{7}{15}\)

\(x=9\frac{7}{15}-7\frac{4}{5}\)

\(x=9\frac{7}{15}-7\frac{12}{15}\)

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

\(x-4\frac{9}{11}=\frac{2121}{2222}\)

\(x-4\frac{9}{11}=\frac{101.21}{101.22}\)

\(x-\frac{53}{11}=\frac{21}{22}\)

\(x=\frac{21}{22}+\frac{53}{11}\)

\(x=\frac{127}{22}=5\frac{17}{22}\)

a) x + \(7\frac{4}{5}\)= \(9\frac{7}{15}\)

x + 39/5 = 95/15

x = 95/15 - 39/5 

x = 95/15 - 127/15

x = -32/15

b) Tương tự

điểm âm của mik chỉ còn 158 nữa thôi, các bn giúp mik nha

Khi nào hết âm mik sẽ ra câu hỏi và giúp lại các bn

Cảm ơn trc~ 

\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)

\(=>x+\dfrac{39}{5}=\dfrac{142}{15}\)

\(=>x=\dfrac{142}{15}-\dfrac{39}{5}=\dfrac{142}{15}-\dfrac{117}{15}\)

\(=>x=\dfrac{25}{15}=\dfrac{5}{3}\)

___________

\(x-4\dfrac{9}{11}=\dfrac{2121}{2222}\)

\(=>x-\dfrac{53}{11}=\dfrac{21}{22}\)

\(=>x=\dfrac{21}{22}+\dfrac{53}{11}=\dfrac{21}{22}+\dfrac{106}{22}\)

\(=>x=\dfrac{127}{22}\)

\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)

\(x=9\dfrac{7}{15}-7\dfrac{4}{5}\)

\(x=\left(9-7\right)+\left(\dfrac{7}{15}-\dfrac{4}{5}\right)\)

\(x=2\dfrac{-1}{3}=\dfrac{5}{3}\)

\(x=\dfrac{2121:101}{2222:101}+4\dfrac{9}{11}\)

\(x=\dfrac{21}{22}+4\dfrac{18}{22}\)

\(x=\dfrac{21+4\cdot22+18}{22}=\dfrac{127}{22}\)

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

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

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)                                 b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

 \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\)                                       \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) 

  \(=\frac{3}{7}.\frac{-9}{9}\)                                                                  \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(=\frac{-3}{7}\)                                                                           \(=\frac{7}{5}-\frac{4}{7}\)

                                                                                               \(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)                                               b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

  \(\frac{-3}{4}x\)           \(=\frac{1}{5}+\frac{4}{10}\)                                     \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\)             \(=\frac{3}{5}\)                                            \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)     

         \(x\)              \(=\frac{3}{5}:\frac{-3}{4}\)                                        \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)                                         

         \(x\)              \(=\frac{4}{-5}\)                                                   \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

                                                                                                             \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) 

                                                                                                                          \(x.\frac{10}{3}=\frac{13}{12}\) 

                                                                                                                                    \(x=\frac{13}{12}:\frac{10}{3}\) 

                                                                                                                                     \(x=\frac{13}{40}\)                             

\(3x.\left(x-\frac{2}{3}\right)=0\)

\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)

\(3x=0\Rightarrow x=0\)

\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)

Vậy.. 

\(A=\left(-\frac{5}{11}\right).\frac{7}{15}+\frac{11}{-5}.\frac{30}{33}\)

\(A=-\frac{7}{33}+-2\)

\(A=-\frac{73}{33}\)

[ A] = -2

làm đc hết rùi phần b thui

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{8}\right)\right]\)

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\frac{5}{8}\right]\)

\(D=\frac{1}{3}-\left(-\frac{15}{8}\right)\)

\(D=\frac{53}{24}\)

\(\left[D\right]=2\)

a) x = 99/20

b) x = 7

c) x = 2

( chỉ lm đc đến đó thui nk )