\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)

\(=\frac{27}{182}-\frac{1}{182}\)

\(=\frac{1}{7}\)

\(\frac{3}{5}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)

\(=\frac{6}{10}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)

\(=\frac{1}{10}.\frac{78}{46}-\frac{1}{10}.\frac{16}{32}\)

\(=\frac{1}{10}\left(\frac{78}{46}-\frac{32}{46}\right)\)

\(=\frac{1}{10}.1=\frac{1}{10}\)

\(=\dfrac{2^{20}\cdot3^{12}+3^{12}\cdot2^{15}}{2^{15}\cdot3^{13}-2^{16}\cdot3^{12}}\)

\(=\dfrac{2^{15}\cdot3^{12}\left(2^5+1\right)}{2^{15}\cdot3^{12}\left(3-2\right)}\)

=33

a, Ta có \(2.3^{x+2}+4.3^{x+1}=3^6.10\)

\(\Rightarrow2.3.3^{x+1}+4.3^{x+1}=3^6.10\)

\(\Rightarrow3^{x+1}.\left(6+4\right)=3^6.10\)

\(\Rightarrow3^{x+1}.10=3^6.10\)

\(\Rightarrow3^{x+1}=3^6\)

\(\Rightarrow x+1=6\)

\(\Rightarrow x=5\)

b,\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{16}\)

\(\Rightarrow\frac{1}{2}.2^{x+4}-2^x=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^{x+3}-2^x=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^x.\left(2^3-1\right)=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^x.\left(2^3-1\right)=-2^{13}.\left(2^3-1\right)\)

\(\Rightarrow2^x=2^{-13}\)

\(\Rightarrow x=-13\)

A ) 2 . 3^x+2 + 4 . 3³⁺¹ = 3⁶ . 10

      2 . 3^x . 9 + 4 . 3^x . 3 = 729 .10

         18 . 3^x + 12 . 3^x = 243 . 3 . 10

                        30 . 3^x = 243 . 30

                               3^x = 243

                                 x = 5 

\(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

Để í ngoặc \(\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]\)

\(\Leftrightarrow\left[\frac{6}{7}+-\frac{6}{7}\right]\)

\(\Leftrightarrow0\)

Vậy biểu thức \(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)có giá trị bằng 0