\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)

2x - 1 = x + 3

2x - x = 3 + 1 

x = 4

2x - 1 =x+3

2x - x = 3+1

x = 4

x + 3 < 6

x < 6 - 3

x < 3

Số dầu tháng này bán được:

10580 : 2= 5290(lít)

Đáp số: 5290 lít dầu

\(A=\dfrac{\sqrt{8}-\sqrt{7}}{8-7}+\sqrt{25\times7}-2\sqrt{2}\\ =\sqrt{8}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}\\ =4\sqrt{7}\)

\(a,A=\left\{x\in N/1+3x;1\le x\le6\right\}\)

\(b,B=\left\{x\in N/x^3;1\le x\le5\right\}\)

a) \(A=\left\{n\inℕ^∗|n=3k+1,k\inℕ,k\le6\right\}\)

b) \(B=\left\{n\inℕ^∗|n=k^3,k\inℕ^∗,k\le5\right\}\)

\(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=\dfrac{3^2.3^8}{3^5.3^3}\)

\(=\dfrac{3^{10}}{3^8}\)

\(=3^2\)

\(=9\)

\(\dfrac{60}{81}< \dfrac{60}{80}\) và \(\dfrac{21}{25}>\dfrac{20}{25}\)

=> \(\dfrac{60}{80}=\dfrac{3}{4}\) và \(\dfrac{20}{25}=\dfrac{4}{5}\) => \(\dfrac{4}{5}>\dfrac{3}{4}\)

\(=>\dfrac{21}{25}>\dfrac{20}{25}>\dfrac{60}{80}>\dfrac{60}{81}\)

Vậy \(\dfrac{21}{25}>\dfrac{60}{81}\)

21/25>60/81