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

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

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

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

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

cái này trong sách í

a) \(\frac{3x-2}{2}=\frac{1-2x}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=2\left(1-2x\right)\)

\(\Leftrightarrow9x-6=2-4x\)

\(\Leftrightarrow13x-8=0\)

\(\Leftrightarrow x=\frac{8}{13}\)

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

\(\Leftrightarrow\frac{4\left(x-1\right)+24}{12}=\frac{36-3\left(2x+5\right)}{12}\)

\(\Leftrightarrow4x-4+24=36-6x-15\)

\(\Leftrightarrow4x+20=-6x+21\)

\(\Leftrightarrow10x=1\)

\(\Leftrightarrow x=\frac{1}{10}\)

c) \(\frac{x-1}{5}+x=\frac{x+1}{7}\)

\(\Leftrightarrow\frac{7\left(x-1\right)+35x}{35}=\frac{5\left(x+1\right)}{7}\)

\(\Leftrightarrow7x-7+35x=5x+5\)

\(\Leftrightarrow37x=12\)

\(\Leftrightarrow x=\frac{12}{37}\)

d) \(2.\left(x-2.5\right)=0,25+\frac{4x-3}{8}\)

\(\Leftrightarrow8\left(x-2,5\right)=2+4x-3\)

\(\Leftrightarrow8x-20=4x-1\)

\(\Leftrightarrow4x=19\)

\(\Leftrightarrow x=\frac{4}{19}\)

a) \(3,6-\left|x-0,4\right|=0\)

\(\Leftrightarrow\left|x-0,4\right|=3,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)

Vậy \(x\in\left\{4;-3,2\right\}\)

b) Ta có:

\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)

Vậy \(x=140\); \(y=70\); \(z=210\)

c)\(\left|x+0,25\right|-4=\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)

d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)

Vậy \(x=\frac{1}{1024}\)

e) \(3^{x-1}+5.3^{x-1}=162\)

\(\Leftrightarrow6.3^{x-1}=162\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

f) \(\frac{x}{-25}=\frac{2}{5}\)

\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)

Vậy \(x=-10\)

g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)

a) \(3,6-\left|x-0,4\right|=0\)

\(\Rightarrow\left|x-0,4\right|=3,6-0\)

\(\Rightarrow\left|x-0,4\right|=3,6.\)

\(\Rightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,6+0,4\\x=\left(-3,6\right)+0,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)

Vậy \(x\in\left\{4;-3,2\right\}.\)

c) \(\left|x+0,25\right|-4=\frac{1}{4}\)

\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{1}{4}+4\)

\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=-\frac{17}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{17}{4}-\frac{1}{4}\\x=\left(-\frac{17}{4}\right)-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{9}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{4;-\frac{9}{2}\right\}.\)

d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)

\(\Rightarrow x:\left(0,25\right)^4=0,25\)

\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)

\(\Rightarrow x=\left(0,25\right)^5\)

\(\Rightarrow x=\frac{1}{1024}\)

Vậy \(x=\frac{1}{1024}.\)

Chúc bạn học tốt!

Câu 1:

a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)

    \(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)

    \(=\frac{1}{2}-\frac{4}{3}\)

     \(=-\frac{5}{6}\)

b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)

    \(=7+\frac{1}{12}+3-\frac{1}{12}-5\)

    \(=5\)

Câu 2:

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)

\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

           Vậy -1\(\le\)x<7

\(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)

\(\Rightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Rightarrow8+4x-10x=5-10x+5\)

\(\Rightarrow8+4x=10\)

\(\Rightarrow4x=2\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy tập nghiệm của phương trình là  A = { 1/2 }

\(\frac{8+4x-10x-5+10x}{20}=0,25\)

\(\frac{4x+3}{20}=0,25\)

\(4x+3=5\)

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

B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)

B=\(2+3\frac{7}{11}\)

B=\(5\frac{7}{11}\)

B = \(5\frac{7}{11}=\frac{62}{11}\)

C = 1

D = \(\frac{5}{2}=2\frac{1}{2}\)

giup mink đi

\(\frac{1}{2}\cdot\left(2-x\right)+0,25=x-\frac{5}{2}:0,4+3\)

\(=>\frac{1}{2}\cdot2-\frac{1}{2}x+\frac{1}{4}=x-\frac{5}{2}:\frac{2}{5}+3\)

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

\(=>\frac{4}{4}+\frac{1}{4}+\frac{25}{4}-\frac{12}{4}=\frac{x}{2}\)

\(=>\frac{18}{4}=\frac{x}{2}\)

\(=>\frac{x}{2}=\frac{9}{2}\)

\(=>x=9\)

Nếu bạn chưa hiểu thì bạn hỏi lại mình nhé! Chúc bạn học tốt!

\(\frac{1}{2}\left(2-x\right)+0,25=x-\frac{5}{2}:0,4+3\)

<=> 1-0,5x+0,25=x-6,25+3

<=> 1,5x=1+0,25+6,25-3=4,5

<=> x=4,5:1,5=3

 vậy x=3

\(\frac{1}{2}\cdot\left(2-x\right)+0,25=x-\frac{5}{2}:0,4+3\)

\(=>\frac{1}{2}\cdot2-\frac{1}{2}\cdot x+\frac{1}{4}=x-\frac{5}{2}:\frac{2}{5}+3\)

\(=>1-\frac{1}{2}x+\frac{1}{4}=x-\frac{5}{2}\cdot\frac{5}{2}+\frac{12}{4}\)

\(=>1+\frac{1}{4}+\frac{25}{4}-\frac{12}{4}=x+\frac{1}{2}x\)

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

\(=>3x=9\)

\(=>x=9:3=3\)

Nếu bạn chưa hiểu thì bạn hỏi lại mình nhé! Chúc bạn học tốt!