\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)

\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)

\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)

\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)

\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)

\(=\dfrac{1}{50}.\dfrac{52}{3}\)

\(=\dfrac{26}{75}\).

Chu vi một cánh hoa:

\(2,8\times3,14=8,792cm\)

Chu vi nụ hoa:

\(2\times3,14=6,28cm\)

Độ dài sợi dây:

\(5\times8,792+6,82=50,24cm\)

Đáp số: 50,24cm.

Đặt \(V=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(\Rightarrow3V=3.\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\right)\)

\(\Rightarrow3V=1+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)

\(\Rightarrow3V=1+V-\dfrac{1}{2187}\)

\(\Rightarrow2V=1-\dfrac{1}{2187}\)

\(\Rightarrow V=\dfrac{1093}{2187}\).

a) 6,28 x 18,24 + 18,24 x 3,72

= (6,28 + 3,72) x 18,24

= 10 x 18,24

= 182,4

b) 35,7 x 99 + 35 + 0,7

= 35,7 x 99 + (35 + 0,7)

= 35,7 x 99 + 35,7

= 35,7 x (99 + 1)

= 35,7 x 100

= 3570

c) 17,34 x 99 + 18 - 0,66

= 17,34 x 99 + (18 - 9,66)

= 17,34 x 99 + 17,34 

= 17,34 x (99 + 1)

= 17,34 x 100

= 1734

d) 0,9 x 95 + 1,8 x 2 + 0,9

= 0,9 x 95 + 0,9 x 2 x 2 + 0,9

= 0,9 x 95 + 0,9 x 4 + 0,9

= 0,9 x (95 + 4 + 1)

= 0,9 x 100

= 90

e) 0,25 x 611,7 x 40

= (0,25 x 40) x 611,7

= 10 x 611,7

= 6117

g) 37,2 x 101 - 37 - 0,2

= 37,2 x 101 - (37 + 0,2)

= 37,2 x 101 - 37,2

= 37,2 x (101 - 1)

= 37,2 x 100

= 3720.

Kẻ \(PQ\perp MN\) ở Q

Theo giả thiết: \(ME=3\times NE\)

\(\Rightarrow NE=\dfrac{1}{4}MN\)

Ta có:

\(S_{\Delta MNP}=\dfrac{1}{2}.PQ.MN\)

\(S_{\Delta NEP}=\dfrac{1}{2}.PQ.NE=\dfrac{1}{2}.PQ.\dfrac{1}{4}.MN=\dfrac{1}{8}PQ.MN\)

\(\Rightarrow\dfrac{S_{\Delta MNP}}{S_{\Delta NEP}}=\dfrac{\dfrac{1}{2}.PQ.MN}{\dfrac{1}{8}.PQ.MN}\)

\(\Leftrightarrow\dfrac{S_{\Delta MNP}}{32,5}=4\)

\(\Leftrightarrow S_{\Delta MNP}=130cm^2\).

Theo đề ra, ta có:

\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Leftrightarrow\left(a^{100}+b^{100}\right).\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)+a^{202}+b^{202}=a^{202}+b^{202}+2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)=2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2-2ab\right)=0\)

\(\Leftrightarrow a=b=0\)

\(\Rightarrow a^{100}+b^{100}=a^{101}+b^{101}\)

\(\Rightarrow a^{100}=a^{101}\)

\(\Leftrightarrow a^{100}.\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow A=a^{2015}+b^{2015}=1+1=2\).