\(\left(X-3\right)\cdot4^5=4^8\\ X-3=4^8:4^5\\ X-3=64\\ X=67\)

2^x+1-1=63

\(2^{x+1}-1=63\\ 2^{x+1}=64\\ 2^{x+1}=2^6\\ =>X+1=6\\ =>x=5\)

\(\left(X-3\right)\cdot4^5=4^8\\ X-3=64\\ =>X=67\)

740:(x+10)=10²-2.13

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

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

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

\(2\cdot3^x-1=53\\ \Leftrightarrow2\cdot3^x=54\\ \Leftrightarrow3^x=27\\ \Leftrightarrow x=3\)

\(2\cdot3^x-1=53\\ 2\cdot3^x=54\\ 3^x=27\\ 3^x=3^3=>x=3\)

...\(\Rightarrow15-6x=3^6:3^5\)

\(\Rightarrow15-6x=3\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

(15-6x).3⁵=3⁶

(15-6x)     =3⁶:3⁵

(15-6x)     =3¹

15-6x       =3

     6x       =15-3

    6x        =12

     x         =12:6

     x         = 2

Vậy x=2

\(...\Rightarrow\left(2x-6\right)=4^9:4^7\)

\(\Rightarrow2x-6=4^2\)

\(\Rightarrow2x-6=16\)

\(\Rightarrow2x=22\)

\(\Rightarrow x=11\)

(2x-6). 4⁷ = 4⁹

2x-6        = \(\dfrac{4^9}{4^7}\)

2x           = 16+6

x             = \(\dfrac{22}{2}\)

x             = 11

75 : (x - 18) = 5²

75 : (x - 18) = 25

         x - 18 = 75 : 25

         x - 18 = 3

                x = 3 + 18 

                x = 21

x=21

\(3+2^{x+1}=35\)

\(\Rightarrow2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.