\(\frac{1}{9}+\frac{8}{9}=\frac{1+8}{9}=\frac{9}{9}=1\)

\(\frac{1}{12}+\frac{2}{12}+\frac{6}{12}+\frac{3}{12}=\frac{1+2+6+3}{12}=\frac{12}{12}=1\)

Chúc bạn học tốt !!

\(\frac{1}{9}\)+\(\frac{8}{9}\)=\(\frac{1+8}{9}\)=\(\frac{9}{9}\)=\(1\)

\(\frac{1}{12}\)+\(\frac{2}{12}\)+\(\frac{6}{12}\)+\(\frac{3}{12}\)=\(\frac{1+2+6+3}{12}\)=\(\frac{12}{12}\)=\(1\)

a.=0

b.= 4

c. = 28

a) ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 ) + ...

= 0 + 0 + 0 + ........

= 0

b ) = -1 + ( -1 ) + ( - 1 ) + ....

= vô tận đây này

c) Cái này vô tận

\(=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)^1}{\left(-5\right)^{20}.3^{10}.3^2}\)\(=\frac{\left(-5\right)^1}{3^2}\)\(=\frac{-5}{9}\)

Hok tốt !

\(\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)^1}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=\frac{-5}{9}\)

\(\frac{ }{ }\)

\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)

\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)

\(\Leftrightarrow x=\frac{-2}{5}\)

\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)

\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)

Đến đây tìm được rồi nhé

3,4, áp dụng bài 1,2 rồi làm :v

lấy máy tính tính 2 vế

xong thay x vào để thỏa mãn điều kiện

hok tốt

Ta có : 

\(\frac{1}{5}+\frac{2}{30}+\frac{121}{165}\le x\le\frac{1}{2}+\frac{156}{72}+\frac{1}{3}\)

\(\Leftrightarrow\)\(\frac{3}{15}+\frac{1}{15}+\frac{11}{15}\le x\le\frac{3}{6}+\frac{13}{6}+\frac{2}{6}\)

\(\Leftrightarrow\)\(\frac{15}{15}\le x\le\frac{18}{6}\)

\(\Leftrightarrow\)\(1\le x\le3\)

\(\Rightarrow\)\(x\in\left\{1;2;3\right\}\)

Vậy \(x\in\left\{1;2;3\right\}\)

Chúc bạn học tốt ~