Question

[Miinhhoa]

Bài 2 Rút gọn: \(A=\dfrac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)

Bài 3:Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng :\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\).

Bài 4 : Cho \(A=3+3^2+3^3+3^4+.....+3^{100}\)Chứng Minh Rằng \(A⋮120\)

giúp mk với nha ngày kia mk nộp rùi !!!!!!!!!

Answer 1

[Akai Haruma]

Bài 2:

\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)

\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)

\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)

\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)

Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)

Thay vào:

\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)

\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)

Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)

Answer 2

[Akai Haruma]

Bài 4:

Ta có:

\(A=3+3^2+3^3+3^4+...+3^{100}\)

\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)

\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)

\(=3.40+3^5.40+....+3^{97}.40\)

\(=120(1+3^4+....+3^{96})\vdots 120\)

Ta có đpcm.