c) \(\frac{3\cdot7-3\cdot19}{3\cdot4}=\frac{3\left(7-19\right)}{3\cdot4}=\frac{3\cdot\left(-12\right)}{3\cdot4}=\frac{3\cdot4\cdot\left(-3\right)}{3\cdot4}=-3\)

Vậy \(\frac{3\cdot7-3\cdot19}{3\cdot4}=-3\)

d,\(\frac{2^3\cdot9^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot\left(3^2\right)^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot3^8-2^4\cdot3^7}{2^2\cdot3^7}=\frac{2^3\cdot3^7\left(3-2\right)}{2^2\cdot3^7}=2\)

bạn làm đúng rồi nhé

chúc bạn học tốt@

Bài 1

a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)

                y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)

                          y= \(\frac{-3}{14}\)

b.5^x + 5^x+2=650

5^x . 1 + 5^x + 5²=650

5^x(1+25)=650

5^x.26=650

5^x=25

x=2

\(S=5+5^2+5^3+5^4+...+5^{2022}\\ =\left(5+5^2\right)+5^2.\left(5+5^2\right)+...+5^{2020}.\left(5+5^2\right)\\ =30+30.5^2+...+30.5^{2020}\\ =30.\left(1+5^2+...+5^{2020}\right)⋮30\)

\(S=5+5^2+5^3+...+5^{2022}\)

\(\Rightarrow S=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{2000}\left(5+5^2\right)\)

\(\Rightarrow S=20+5^2.20+...+5^{2000}.20\)

\(\Rightarrow S=20\left(1+5^2+...+5^{2000}\right)⋮20\)

\(\Rightarrow dpcm\)

Giải:

a) \(\left(-\dfrac{5}{28}+1,75+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)

\(=\left(-\dfrac{5}{28}+\dfrac{7}{4}+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)

\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-\dfrac{39}{20}\right)\)

\(=\dfrac{9}{5}:\left(-\dfrac{39}{20}\right)\)

\(=\dfrac{9.\left(-20\right)}{5.39}\)

\(=\dfrac{3.\left(-4\right)}{1.13}\)

\(=\dfrac{-12}{13}\)

b) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{22}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{22}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

c) \(70,5-528:\dfrac{15}{2}\)

\(=70,5-528.\dfrac{2}{15}\)

\(=70,5-\dfrac{1056}{15}\)

\(=70,5-70,4\)

\(=0,1\)

a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)

\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)

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

Vậy \(x=-\dfrac{1}{2}\).

b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)

\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)

\(3x-3\cdot7=-\dfrac{19}{2}\)

\(3x-21=-\dfrac{19}{2}\)

\(3x=-\dfrac{19}{2}+21\)

\(3x=\dfrac{23}{2}\)

\(x=\)\(\dfrac{23}{2}:3\)

\(x=\dfrac{23}{6}\)

Vậy \(x=\dfrac{23}{6}\).

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)

\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)

\(\dfrac{3}{4x}=-\dfrac{65}{22}\)

\(4x=\dfrac{3\cdot22}{-65}\)

\(4x=-\dfrac{66}{65}\)

\(x=-\dfrac{66}{65}:4\)

\(x=-\dfrac{33}{130}\)

Vậy \(x=-\dfrac{33}{130}\).

d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)

\(\left|x\right|=\dfrac{29}{12}\)

\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)

Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).

n²-3n+13\(⋮\)n+3

n²-3(n+3)+4\(⋮\)n+3

4\(⋮\)n+3

n+3\(\in\)Ư(4)=\(\left\{\pm1;\pm2;\pm4\right\}\)