a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

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

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

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

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

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

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

a, => x^3 < 0 ; x-3 > 0 hoặc x^3 > 0 ; x-3 < 0

=> 0 < x < 3

b, => x^4.(2x-8) < 0

=> x^4.(x-4) < 0

Vì x^4 >= 0

=> x-4 < 0

=> x  < 4

c, Vì x-1 < x+12

=> x-1 < 0 ; x+12 >0

=> -12 < x < 1

d, => x-12 > 0 ; x-1 > 0 hoặc x-12 < 0 ; x-1 < 0

=> x  >12 hoặc x < 1

Tk mk nha

Thank you so much

\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)

\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)

=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)

\(\Rightarrow7x< 36< 7x+7\)

\(\Rightarrow x< \dfrac{36}{7}< x+1\)

\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)

\(\Rightarrow x=5\)

tik cho mình nhé

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a.

\(\%_{Fe_{\left(Fe_2O_3\right)}}=\dfrac{56.2}{160}.100\%=70\%\)

\(\%_{O_{\left(Fe_2O_3\right)}}=100\%-70\%=30\%\)

b.

\(\%_{C_{\left(C_6H_{12}O_6\right)}}=\dfrac{12.6}{180}.100\%=7\%\)

\(\%_{H_{\left(C_6H_{12}O_6\right)}}=\dfrac{1.12}{180}.100\%=6,7\%\)

\(\%_{O_{\left(C_6H_{12}O_6\right)}}=100\%-7\%-6,7\%=86,3\%\)

c.

\(\%_{C_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{12.6}{162n}.100\%=44,4n\%\)

\(\%_{H_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{1.10}{162n}.100\%=6,2n\%\)

\(\%_{O_{\left(C_6H_{1o}O_5\right)}}=\dfrac{16.5}{162n}.100\%=49,4n\%\)

\(\Rightarrow49,4n\%=100\%-44,4n\%-6,2n\%\)

\(\Leftrightarrow n=1\)

\(\Rightarrow\left\{{}\begin{matrix}\%_C=44,4\%\\\%_H=6,2\%\\\%_O=49,4\%\end{matrix}\right.\)

d. 

\(\%_{Na_{\left(NaCl\right)}}=\dfrac{23}{58,5}.100\%=39,3\%\)

\(\%_{Cl_{\left(NaCl\right)}}=100\%-39,3\%=60,7\%\)