\(D=5^7+5^{11}+5^{15}+...+5^{4k+3}\)   (1)

Nhân D với 5⁴ ta được:

\(5^4.D=5^{11}+5^{15}+...+5^{4k+7}\)      (2)

Lấy (2) trừ đi (1) ta được:

    \(5^4.D-D=5^{4k+7}-5^7\)

=> \(D=\frac{5^7\left(5^{4k}-1\right)}{5^4-1}\)

Thanks co a !

\(3x^2-2x.\left(5+1,5x\right)+10\)

\(=3x^2-2x.5-2x.1,5x+10\)

\(=3x^2-3x^2-10x+10\)

\(=10-10x\)

\(=10.\left(1-x\right)\)

cam on ban nha

15/24 = 5/8 ; 21/24 = 7/8 ; 54/18 = 9/3 ; 100/25 = 4/1.

15/24=5/8;21/42=1/2;54/18=3/1;100/25=4/1

bn ơi ghi lại đề bài = phân số đc hk ghi cả nhân hay chia nx khó nhìn lắm!

ghi the duoc roi

\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)

Đặt \(B=2^{99}+2^{98}+...+2+1\)

\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)

\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)

\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)

:(

:)

\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)

Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B

\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)

Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C

\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)

\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)

\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)