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\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=\frac{17.2}{7.13}+\frac{17.3}{13.22}+\frac{17.5}{22.37}+\frac{17.4}{37.49}\)
\(=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+...+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)\(=\frac{17}{3}.\frac{6}{49}=\frac{17.2}{49}=\frac{34}{49}\)
Có : \(\frac{17}{24}=\frac{34}{48}\)
Vì 48 < 49 => \(\frac{34}{48}>\frac{34}{49}\). Hay \(\frac{17}{24}>A\)
1 - 4 - 7 + 10 + 13 - 16 - 19 + 22 + 25 - 28 - 31 + 34 - 37
= ( 1 - 4 - 7 + 10 ) + ( 13 - 16 - 19 + 22 ) + ( 25 - 28 - 31 + 34 ) - 37
= 0 + 0 + 0 - 37
= -37
1 - 4 - 7 + 10 + 13 -16 - 19 + 22- 25 + 28 - 31 - 34 - 37 = - 99
\(a)\) \(5^{13}:5^{10}-25.2^2=5^3-25.4=125-100=25\)
\(b)\) \(20:2^2+5^9:5^8=20:4+5^1=5+5=10\)
\(c)\) \(100:5^2+7.3^2=100:25+7.9=4+36=40\)
\(d)\) \(84:4+3^9:3^7+5^0=84:4+3^2+1=21+9+1=31\)
\(e)\)
\(29-\left[16+3.\left(51-49\right)\right]=29-\left[16+3.2\right]=29-\left[16+6\right]=29-22=7\)
\(f)\) \(5.2^2+98:7^2=5.4+98:49=20+2=22\)
\(g)\) \(3^{11}:3^9-147:7^2=3^2-147:49=9-3=6\)
\(295-\left(31-2^2.5\right)^2=295-\left(31-4.5\right)^2=295-\left(31-20\right)^2=295-11^2=295-121=174\)
\(7^{18}:7^{16}+2^2.3^2=7^2+4.9=49+36=85\)
a: \(5^{13}:5^{10}-25\cdot2^2\)
\(=5^3-25\cdot4\)
=125-100
=25
b: \(20:2^2+5^9:5^8\)
\(=20:4+5\)
=5+5
=10