\(=\frac{-26}{9}\)

a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)

\(=\frac{11651}{720}\)

B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)

\(=\frac{10807}{810}\)

C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)

\(=\frac{2893}{200}\)

CHÚC BN HỌC TỐT!!!!!
 

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

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

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

B=\(\frac{1}{3}+\frac{-0,175+\frac{5}{11}}{0,2625-\frac{15}{22}}\)

B=\(\frac{1}{3}+\frac{\frac{123}{440}}{-\frac{369}{880}}\)

B=\(0,2161148416\)

 Thay x vào ta có: 

\(B=\frac{1}{3}+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)

\(B=\frac{1}{3}+\frac{\frac{2}{10}-\frac{375}{1000}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(B=\frac{1}{3}+\frac{\frac{1}{5}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(B=\frac{1}{3}+\frac{\frac{123}{440}}{\frac{-369}{880}}\)

\(B=\frac{1}{3}+\frac{123}{440}.\frac{880}{-369}\)

\(B=\frac{1}{3}+\frac{-2}{3}\)

\(B=\frac{-1}{3}\)

Vậy...

a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)

\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)

\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)

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

hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)

Vậy: \(x=\frac{4}{3}\)

b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)

\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)

hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)

Vậy: \(x=\frac{13}{15}\)

c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)

\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)

\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)

\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)

\(\Leftrightarrow3x=-160\)

hay \(x=\frac{-160}{3}\)

Vậy: \(x=\frac{-160}{3}\)

d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)

a/ (x - 2)³ + \(\frac{8}{27}\) = 0

=> (x - 2)³ = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)

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

=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)

b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)

=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)

=> \(x=\frac{13}{60}.4=\frac{13}{15}\)

c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)

=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)

=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)

=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)

=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)

nhầm sửa lạ nha 

tử bằng 1

 mẫu bằng 

(1/2-1/5+1/4-1/5):(3/10-1/5)=7/20:1/10=7/2

vậy 1:7/2=2/7