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1) \(125^5:25^7\)
\(=\left(5^3\right)^5:\left(5^2\right)^7\)
\(=5^{15}:5^{14}\)
= 5
2) \(27^8:9^9\)
\(=\left(3^3\right)^8:\left(3^2\right)^9\)
\(=3^{24}:3^{18}\)
\(=3^6\)
3) \(36^5:6^8\)
\(=\left(6^2\right)^5:6^8\)
\(=6^{10}:6^8\)
\(=6^2\)
4) \(49^6:7^{10}\)
\(=\left(7^2\right)^6:7^{10}\)
\(=7^{12}:7^{10}=7^2\)
5) \(7^{20}:49^9\)
\(=7^{20}:\left(7^2\right)^9\)
\(=7^{20}:7^{18}=7^2\)
6) \(\frac{1}{2^{10}}:\frac{1}{8^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{\left(2^3\right)^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{2^9}=\frac{1}{2^{10}}.\frac{2^9}{1}=\frac{1}{2}\)
7) \(\left(-\frac{1}{2}\right)^{21}:\frac{1}{4^{10}}\)
\(=\frac{\left(-1\right)^{21}}{2^{21}}:\frac{1}{\left(2^2\right)^{10}}\)
\(=-\frac{1}{2^{21}}:\frac{1}{2^{20}}=-\frac{1}{2^{21}}.\frac{2^{20}}{1}\)
\(=-\frac{1}{2}\)
8) \(\frac{1}{16^5}:\left(-\frac{1}{2}\right)^{18}\)
\(=\frac{1}{\left(2^4\right)^5}:\frac{\left(-1\right)^{18}}{2^{18}}\)
\(=\frac{1}{2^{20}}:\frac{1}{2^{18}}\)
\(=\frac{1}{2^{20}}.\frac{2^{18}}{1}=\frac{1}{4}\)
9) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
Ta có quy luật:
1+1=12 = 1
2+2=22 = 4
3+3 = 32 = 9
..................
.......................
9 + 9 = 92 = 81
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo)
a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha
.
\(\left(1+2+3+...+100\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}\right)\cdot\dfrac{\left(1+8+27+64+125+216+343\right)}{1+4+9+16+25+36+49}\)
\(=100\cdot\dfrac{101}{2}\cdot\left(\dfrac{12}{24}+\dfrac{6}{24}+\dfrac{4}{24}+\dfrac{3}{24}\right)\cdot\dfrac{\left(1+2+3+4+5+6+7\right)^2}{140}\)
\(=101\cdot50\cdot\dfrac{25}{24}\cdot\dfrac{784}{140}\)
\(=5050\cdot\dfrac{35}{6}=\dfrac{88375}{3}\)