Đặt \(B=4^{1993}+4^{1992}+.......+4^2+1\)

\(\Rightarrow4B=4^{1994}+4^{1993}+....+4^3+4\)

\(\Rightarrow3B=4^{1994}-1\)

Mà: \(A=75B+25=25\left(3B+1\right)=25\left(4^{1994}-1+1\right)=25.4^{1994}\)

\(A=75\left(4^{1993}+5^{1992}+...+4^2+5\right)+25=75B+25\)

Xét \(B=4^{1993}+4^{1992}+...+4^2+5=4^{1993}+4^{1992}+...+4^2+4+1\)

\(\Rightarrow4B=4^{1994}+4^{1993}+...+4^2+4\)

\(\Rightarrow4B+1-4^{1994}=4^{1993}+4^{1992}+...+4^2+4+1=B\)

\(\Rightarrow3B=4^{1994}-1\Rightarrow B=\dfrac{4^{1994}-1}{3}\)

Vậy \(A=75.\dfrac{\left(4^{1994}-1\right)}{3}+25=25.4^{1994}-25+25\)

\(\Rightarrow A=25.4^{1994}\)

Ta có \(A=75\left(4^{1993}+4^{1992}+....+4+1\right)+25\)

\(\Leftrightarrow A=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)+25\)

Vận dụng hằng đẳng thức

\(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+b^{n-1}\right)\)

Ta có

\(\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)=4^{1994}-1\)

\(\Rightarrow A=25\left(4^{1994}-1\right)+25\)

\(\Leftrightarrow A=25\cdot4^{1994}\)

Vậy \(A=25\cdot4^{1994}\)

\(\dfrac{x-1}{1992}+\dfrac{x-2}{1993}=\dfrac{x-3}{1994}+\dfrac{x-4}{1995}\)

\(\Rightarrow\left(\dfrac{x-1}{1992}+1\right)+\left(\dfrac{x-2}{1993}+1\right)=\left(\dfrac{x-3}{1994}+1\right)+\left(\dfrac{x-4}{1995}+1\right)\)

\(\Rightarrow\left(\dfrac{x-1+1992}{1992}\right)+\left(\dfrac{x-2+1993}{1993}\right)=\left(\dfrac{x-3+1994}{1994}\right)+\left(\dfrac{x-4+1995}{1995}\right)\)

\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}=\dfrac{x+1991}{1994}+\dfrac{x+1991}{1995}\)

\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}-\dfrac{x+1991}{1994}-\dfrac{x+1991}{1995}=0\)

\(\Rightarrow\left(x+1991\right)\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)=0\)

\(\Rightarrow\left(x+1991\right)=0\) ( vì \(\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)\ne0\)

\(\Rightarrow x=-1991\)

\(A=75\left(4^{1993}+4^{1992}+...+4^2+5\right)+31\)

\(=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4^2+4+1\right)+31\)

\(=25\left(4^{1994}+4^{1993}+...+4^3+4^2+4-4^{1993}-....-4-1\right)+31\)

\(=25.\left(4^{1994}-1\right)+31\)

\(=25.4^{1994}-25+31\)

\(=25.4^{1994}+6\)

                                                              Bài giải

\(A=75\cdot\left(4^{1993}+4^{1992}+...+4^2+4\right)+31\)

Đặt \(B=4^{1993}+4^{1992}+...+4^2+4\)

\(B=4+4^2+...+4^{1992}+4^{1993}\)

\(4B=4^2+4^3+...+4^{1993}+4^{1994}\)

\(4B-B=3B=4^{1994}-4\)

\(B=\frac{4^{1994}-4}{3}\)

Thay \(B=\frac{4^{1994}-4}{3}\) vào biểu thức ta có : 

\(A=75\cdot\frac{4^{1994}-4}{3}+31\)

\(B=25\cdot3\cdot\frac{4^{1994}-4}{3}+31\)

\(B=25\cdot\left(4^{1994}-4\right)+31\)

=> B = 75.4¹⁹⁹³ + 75.4¹⁹⁹² + ... + 75.4 + 75 + 25

        = 25.3.4.4¹⁹⁹² + 25.3.4.4¹⁹⁹¹ + ... + 25.3.4 + 100

        = 100.3.4¹⁹⁹² + 100.3.4¹⁹⁹¹ + ... + 100.3 + 100

        = 100 ( 4¹⁹⁹² + 4¹⁹⁹¹ + .... + 3 + 1 ) CHIA HẾT CHO 100

vậy cho mình hỏi Đinh Đức Hùng, số 4¹⁹⁹³ sẽ sao ạ ?

Mình khai chiển 4¹⁹⁹³ = 4.4¹⁹⁹² rồi như , bạn chưa nhìn ra àk