Yen Nhi
Giới thiệu về bản thân
\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)
\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)
\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)
\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)
\(=\dfrac{1}{50}.\dfrac{52}{3}\)
\(=\dfrac{26}{75}\).
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:
\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Leftrightarrow\left(a^{100}+b^{100}\right).\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)+a^{202}+b^{202}=a^{202}+b^{202}+2a^{101}.b^{101}\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)=2a^{101}.b^{101}\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2-2ab\right)=0\)
\(\Leftrightarrow a=b=0\)
\(\Rightarrow a^{100}+b^{100}=a^{101}+b^{101}\)
\(\Rightarrow a^{100}=a^{101}\)
\(\Leftrightarrow a^{100}.\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)
\(\Rightarrow A=a^{2015}+b^{2015}=1+1=2\).