1) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{7}{8}+\dfrac{1}{3}\\\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{24}\\\dfrac{1}{2}x=\dfrac{29}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-\dfrac{13}{24}\right):\dfrac{1}{2}\\x=\dfrac{29}{24}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)

2) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)

\(\dfrac{1}{12}\). \(\dfrac{37}{39}+\dfrac{1}{12}.\dfrac{2}{39}+\dfrac{1}{4}\)

=\(\dfrac{1}{12}.\left(\dfrac{37}{39}+\dfrac{2}{39}\right)+\dfrac{1}{4}\)

=\(\dfrac{1}{12}.1+\dfrac{1}{4}\)

=\(\dfrac{13}{12}+\dfrac{1}{4}\)

=\(\dfrac{16}{12}\)

d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)

h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)

\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)

k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)

\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)

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

n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)

\(=3-1+\dfrac{1}{8}\cdot16\)

=2+2

=4

\(=5+\dfrac{17}{39}+4+\dfrac{2}{85}-3-\dfrac{2}{45}-13-\dfrac{19}{85}+\dfrac{11}{45}-\dfrac{4}{39}\)

\(=\left(-4\right)+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}=-4+\dfrac{1}{3}=-\dfrac{11}{3}\)

2/5 . x = (1/2)³ : 1/2

2/5 . x = 1/8 : 1/2

2/5 . x = 1/4

x = 1/4 : 2/5

x = 5/8

\(\Leftrightarrow\dfrac{2}{5}x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{5}{2}=\dfrac{5}{8}\)

a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)

b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)

c)Tương tự câu b, ta đc:

\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)

d)Tương tự câu a, ta đc:

\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)

Chúc Bạn Học Tốt!!!

a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)

b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)

c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)

d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)

\(A=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)

\(A=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)

\(A=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+...+\dfrac{1}{9^2.10}\)

\(A=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+....+\dfrac{1}{900}\)

\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{3}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\........\\\dfrac{1}{900}< \dfrac{1}{32}\end{matrix}\right.\)

Nên \(A< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{32}.8=1\)