Bài 4:
a: Đặt \(C=3^{2019}+3^{2018}+\cdots+3^2+3+1\)
=>\(3C=3^{2020}+3^{2019}+\cdots+3^3+3^2+3\)
=>3C-C=\(3^{2020}+3^{2019}+\cdots+3-3^{2019}-3^{2018}-\cdots-1\)
=>2C=\(3^{2020}-1\)
=>\(C=\frac{3^{2020}-1}{2}\)
Ta có: \(A=3^{2020}-\left(3^{2019}+3^{2018}+\cdots+3^2+3+1\right)\)
\(=3^{2020}-\frac{3^{2020}-1}{2}=\frac{2\cdot3^{2020}-3^{2020}+1}{2}=\frac{3^{2020}+1}{2}\)
b: Đặt \(D=5^{2009}+5^{2008}+\cdots+5^1+5^0\)
=>5D=\(5^{2010}+5^{2009}+\cdots+5\)
=>5D-D=\(5^{2010}+5^{2009}+\cdots+5-5^{2009}-5^{2008}-\cdots-5-1\)
=>4D=\(5^{2010}-1\)
=>\(D=\frac{5^{2010}-1}{4}\)
Ta có: \(B=5^{2010}-\left(5^{2009}+5^{2008}+\cdots+5+1\right)\)
\(=5^{2010}-\frac{5^{2010}-1}{4}=\frac{4\cdot5^{2010}-5^{2010}+1}{4}=\frac{3\cdot5^{2010}+1}{4}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\cdot\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
