1: \(2x+\dfrac{1}{2}=-\dfrac{5}{3}\)

=>\(2x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{10}{6}-\dfrac{3}{6}=-\dfrac{13}{6}\)

=>\(x=-\dfrac{13}{6}:2=-\dfrac{13}{12}\)

2: \(\dfrac{1}{7}-\dfrac{3}{5}x=\dfrac{3}{5}\)

=>\(\dfrac{3}{5}x=\dfrac{1}{7}-\dfrac{3}{5}=\dfrac{5}{35}-\dfrac{21}{35}=-\dfrac{16}{35}\)

=>\(x=-\dfrac{16}{35}:\dfrac{3}{5}=-\dfrac{16}{35}\cdot\dfrac{5}{3}=\dfrac{-16}{7\cdot3}=-\dfrac{16}{21}\)

3: \(-3x-\dfrac{3}{4}=\dfrac{6}{5}\)

=>\(-3x=\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{39}{20}\)

=>\(x=\dfrac{39}{20}:\left(-3\right)=-\dfrac{13}{20}\)

4: \(\dfrac{3}{7}-\dfrac{1}{2}x=\dfrac{5}{3}\)

=>\(\dfrac{1}{2}x=\dfrac{3}{7}-\dfrac{5}{3}=\dfrac{9}{21}-\dfrac{35}{21}=-\dfrac{26}{21}\)

=>\(x=-\dfrac{26}{21}\cdot2=-\dfrac{52}{21}\)

5: \(2x-\dfrac{3}{4}=-\dfrac{5}{8}\)

=>\(2x=-\dfrac{5}{8}+\dfrac{3}{4}=-\dfrac{5}{8}+\dfrac{6}{8}=\dfrac{1}{8}\)

=>\(x=\dfrac{1}{8}:2=\dfrac{1}{16}\)

6: \(\dfrac{3}{5}-\dfrac{1}{4}x=\dfrac{2}{5}\)

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

=>\(x=\dfrac{1}{5}:\dfrac{1}{4}=\dfrac{4}{5}\)

7: \(-\dfrac{2}{3}x+2=\dfrac{3}{4}\)

=>\(-\dfrac{2}{3}x=\dfrac{3}{4}-2=-\dfrac{5}{4}\)

=>\(x=\dfrac{5}{4}:\dfrac{2}{3}=\dfrac{5}{4}\cdot\dfrac{3}{2}=\dfrac{15}{8}\)

8: \(\dfrac{2}{5}x+\dfrac{3}{2}=\dfrac{5}{4}\)

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

=>\(x=-\dfrac{1}{4}:\dfrac{2}{5}=-\dfrac{1}{4}\cdot\dfrac{5}{2}=-\dfrac{5}{8}\)

9: \(\dfrac{3}{4}x-\dfrac{5}{2}=\dfrac{4}{3}\)

=>\(\dfrac{3}{4}x=\dfrac{5}{2}+\dfrac{4}{3}=\dfrac{15}{6}+\dfrac{8}{6}=\dfrac{23}{6}\)

=>\(x=\dfrac{23}{6}:\dfrac{3}{4}=\dfrac{23}{6}\cdot\dfrac{4}{3}=\dfrac{92}{18}=\dfrac{46}{9}\)

10: \(-2x+\dfrac{3}{14}=\dfrac{1}{7}-\dfrac{4}{21}\)

=>\(-2x+\dfrac{3}{14}=\dfrac{3}{21}-\dfrac{4}{21}=-\dfrac{1}{21}\)

=>\(-2x=-\dfrac{1}{21}-\dfrac{3}{14}=-\dfrac{2}{42}-\dfrac{9}{42}=-\dfrac{11}{42}\)

=>\(x=\dfrac{11}{84}\)

11: \(\dfrac{3}{2}x-\dfrac{1}{4}=\dfrac{2}{3}\)

=>\(\dfrac{3}{2}x=\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{8}{12}+\dfrac{3}{12}=\dfrac{11}{12}\)

=>\(x=\dfrac{11}{12}:\dfrac{3}{2}=\dfrac{11}{12}\cdot\dfrac{2}{3}=\dfrac{22}{36}=\dfrac{11}{18}\)

12: \(\dfrac{2}{3}-\dfrac{4}{7}x=1-\dfrac{1}{2}\)

=>\(\dfrac{2}{3}-\dfrac{4}{7}x=\dfrac{1}{2}\)

=>\(\dfrac{4}{7}x=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}\)

=>\(x=\dfrac{1}{6}:\dfrac{4}{7}=\dfrac{1}{6}\cdot\dfrac{7}{4}=\dfrac{7}{24}\)

13: \(-3x+\dfrac{1}{2}=2-\dfrac{4}{3}\)

=>\(-3x+\dfrac{1}{2}=\dfrac{6}{3}-\dfrac{4}{3}=\dfrac{2}{3}\)

=>\(-3x=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\)

=>\(x=\dfrac{1}{6}:\left(-3\right)=-\dfrac{1}{18}\)

14: \(\dfrac{3}{4}-\dfrac{5}{6}x=2-\dfrac{2}{3}\)

=>\(\dfrac{3}{4}-\dfrac{5}{6}x=\dfrac{4}{3}\)

=>\(\dfrac{5}{6}x=\dfrac{3}{4}-\dfrac{4}{3}=\dfrac{9}{12}-\dfrac{16}{12}=-\dfrac{7}{12}\)

=>\(x=-\dfrac{7}{12}:\dfrac{5}{6}=-\dfrac{7}{12}\cdot\dfrac{6}{5}=-\dfrac{7}{2\cdot5}=-\dfrac{7}{10}\)

1: \(C=\dfrac{5}{18}+\dfrac{8}{19}-\dfrac{7}{21}+\left(-\dfrac{10}{36}+\dfrac{11}{19}+\dfrac{1}{3}\right)-\dfrac{5}{8}\)

\(=\dfrac{5}{18}+\dfrac{8}{19}-\dfrac{1}{3}-\dfrac{5}{18}+\dfrac{11}{19}+\dfrac{1}{3}-\dfrac{5}{8}\)

\(=\left(\dfrac{5}{18}-\dfrac{5}{18}\right)+\left(\dfrac{8}{19}+\dfrac{11}{19}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)-\dfrac{5}{8}\)

\(=1-\dfrac{5}{8}=\dfrac{3}{8}\)

2: \(E=\dfrac{1}{13}+\left(-\dfrac{5}{18}-\dfrac{1}{13}+\dfrac{12}{17}\right)-\left(\dfrac{12}{17}-\dfrac{5}{18}+\dfrac{7}{5}\right)\)

\(=\dfrac{1}{13}+\dfrac{-5}{18}-\dfrac{1}{13}+\dfrac{12}{17}-\dfrac{12}{17}+\dfrac{5}{18}-\dfrac{7}{5}\)

\(=\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(-\dfrac{5}{18}+\dfrac{5}{18}\right)+\left(\dfrac{12}{17}-\dfrac{12}{17}\right)-\dfrac{7}{5}\)

\(=-\dfrac{7}{5}\)

3: \(F=\dfrac{15}{14}-\left(\dfrac{17}{23}-\dfrac{80}{87}+\dfrac{5}{4}\right)+\left(\dfrac{17}{23}-\dfrac{15}{14}+\dfrac{1}{4}\right)\)

\(=\dfrac{15}{14}-\dfrac{17}{23}+\dfrac{80}{87}-\dfrac{5}{4}+\dfrac{17}{23}-\dfrac{15}{14}+\dfrac{1}{4}\)

\(=\left(\dfrac{15}{14}-\dfrac{15}{14}\right)+\left(-\dfrac{17}{23}+\dfrac{17}{23}\right)+\dfrac{80}{87}-1\)

\(=\dfrac{80}{87}-1=-\dfrac{7}{87}\)

4: \(G=\dfrac{1}{25}-\dfrac{4}{27}+\left(-\dfrac{23}{27}+\dfrac{-1}{25}-\dfrac{5}{43}\right)+\dfrac{5}{43}-\dfrac{4}{7}\)

\(=\dfrac{1}{25}-\dfrac{4}{27}-\dfrac{23}{27}-\dfrac{1}{25}-\dfrac{5}{43}+\dfrac{5}{43}-\dfrac{4}{7}\)

\(=\left(\dfrac{1}{25}-\dfrac{1}{25}\right)+\left(-\dfrac{4}{27}-\dfrac{23}{27}\right)+\left(-\dfrac{5}{43}+\dfrac{5}{43}\right)-\dfrac{4}{7}\)

\(=-1-\dfrac{4}{7}=-\dfrac{11}{7}\)

5: \(H=\dfrac{4}{15}-\dfrac{23}{28}\left(-\dfrac{23}{28}+\dfrac{-11}{15}-\dfrac{29}{27}\right)-\dfrac{2}{27}\)

\(=\dfrac{4}{15}-\dfrac{23}{28}\cdot\left(-\dfrac{23}{28}+\dfrac{-99}{135}-\dfrac{145}{135}\right)-\dfrac{2}{27}\)

\(=\dfrac{4}{15}-\dfrac{2}{27}+\dfrac{23}{28}\cdot\dfrac{23}{28}+\dfrac{23}{28}\cdot\dfrac{244}{135}\)

\(=\dfrac{26}{135}+\dfrac{529}{784}+\dfrac{1403}{945}=\dfrac{1585}{945}+\dfrac{529}{784}\)

\(=\dfrac{317}{189}+\dfrac{529}{784}=\dfrac{317\cdot784+529\cdot189}{189\cdot784}\)

\(=\dfrac{348509}{148176}\)

6: \(K=\dfrac{1}{16}-\dfrac{5}{21}+\left(-\dfrac{1}{16}+\dfrac{-3}{5}-\dfrac{-5}{21}\right)+\dfrac{-2}{5}+\dfrac{3}{4}\)

\(=\dfrac{1}{16}-\dfrac{5}{21}-\dfrac{1}{16}-\dfrac{3}{5}+\dfrac{5}{21}+\dfrac{-2}{5}+\dfrac{3}{4}\)

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

1: \(\dfrac{3}{7}+\left(-\dfrac{5}{2}\right)+\left(-\dfrac{3}{5}\right)\)

\(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}\)

\(=\dfrac{30}{70}-\dfrac{175}{70}-\dfrac{42}{70}\)

\(=-\dfrac{187}{70}\)

2: \(\left(-\dfrac{4}{3}\right)+\left(-\dfrac{2}{5}\right)+\left(-\dfrac{3}{2}\right)\)

\(=-\dfrac{40}{30}+\dfrac{-12}{30}+\dfrac{-45}{30}\)

\(=\dfrac{-40-12-45}{30}=\dfrac{-97}{30}\)

3: \(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\)

\(=\dfrac{8}{10}-\dfrac{7}{10}+\dfrac{2}{7}\)

\(=\dfrac{1}{10}+\dfrac{2}{7}=\dfrac{7}{70}+\dfrac{20}{70}=\dfrac{27}{70}\)

4: \(\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}+\dfrac{7}{4}+\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{14}{8}+\dfrac{4}{8}+\dfrac{3}{8}\)

\(=\dfrac{2}{3}+\dfrac{21}{8}=\dfrac{16}{24}+\dfrac{63}{24}=\dfrac{79}{24}\)

a: Xét ΔABC vuông tại A có \(BC^2=AB^2+AC^2\)

=>\(BC=\sqrt{24^2+18^2}=30\left(cm\right)\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=AB^2\\CH\cdot CB=CA^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{24^2}{30}=19,2\left(cm\right)\\CH=\dfrac{18^2}{30}=10,8\left(cm\right)\end{matrix}\right.\)

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}=\dfrac{18}{30}=\dfrac{3}{5}\)

nên \(\widehat{B}\simeq37^0\)

b: Xét ΔAHC vuông tại H có \(AH^2+HC^2+AC^2\)

=>\(AH^2=AC^2-HC^2\left(1\right)\)

Xét ΔAHB vuông tại H có HM là đường cao

nên \(AH^2=AM\cdot AB\left(2\right)\)

Từ (1),(2) suy ra \(AM\cdot AB=AC^2-HC^2\)

\(N=\dfrac{\sqrt{x}+1}{5}\in N\)

\(\Rightarrow\sqrt{x}+1⋮5\)

\(\Rightarrow\sqrt{x}+1\in B\left(5\right)=\left\{5;10;15;20;...\right\}\)

\(\Rightarrow x\in\left\{16;81...\right\}\)

mà \(x< 29\)

\(\Rightarrow x\in\left\{16\right\}\)

\(M=1-\left|x+3\right|-\left|y-5\right|\)

\(\Rightarrow M=1-\left(\left|x+3\right|+\left|y-5\right|\right)\)

mà \(\left|x+3\right|;\left|y-5\right|\ge0,\forall x;y\in R\)

\(\Rightarrow M=1-\left(\left|x+3\right|+\left|y-5\right|\right)\le1\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+3=0\\y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=5\end{matrix}\right.\)

Vậy \(GTLN\left(M\right)=1\left(tại.x=-3;y=5\right)\)

a) 

 vuông cân 

 có : 

 là góc chung

\(\widehat{D}=\widehat{A}=90^o\)

\(⇒ΔCED\simΔCBA( g.g)\) 

Xét : 

\(\widehat{C}\) là góc chung

\(\RightarrowΔCAD\simΔCBE\) 

\(\widehat{BEC}=\widehat{ADC}\)

Tacó \(\widehat{ADC}+\widehat{HDA}=180^o\)

mà 

 vuông cân

\(\Rightarrow AB=AE\)

b) Ta có :

\(\widehat{AMB}=\widehat{AHB}=90^o\)

Nên tứ giác \(AMHB\) là tứ giác nội tiếp đường tròn đường kính \(AB\)

\(\Rightarrow\widehat{AHM}=\widehat{ABM}=45^o\)