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Đặt N=2^49+2^48+...+2^3+2^2+2
=>2N=2^50+2^49+...+2^4+2^3+2^2
=>2N-N=2^50-2
=>N=2^50-2
=>M=2^50-2^50+2=2
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)
c) Đề hơi sai roi bạn oi
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)
a) 25^2:5^2 = 25
nếu đề là 25^3:5^2 thì :
=(5^2)^3:5^2
=5^6:5^2
=5^4
b) =(3/7)^21:[(3/7)^2] ^6
=(3/7)^21:[(3/7)^2]
=(3/7)21-12
=3/7^9
\(25^3\div5^2=\left(5^2\right)^3\div5^2=5^6\div5^2=5^4=625\)
\(\left(\frac{3}{7}\right)^{21}\div\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}\div\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}\div\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2\div2=3-1+\frac{1}{4}\times\frac{1}{2}=2+\frac{1}{8}=\frac{17}{8}\)
\(Tacó\frac{8^{19}\times14^{11}}{49^6\times2^{68}}\)
\(=\frac{\left(2^3\right)^{19}\times2^{11}\times7^{11}}{\left(7^2\right)^6\times2^{68}}\)
\(=\frac{2^{57}\times2^{11}\times7^{11}}{7^{12}\times2^{68}}\)
\(=\frac{2^{68}\times7^{11}}{7^{12}\times2^{68}}\)
\(=\frac{1\times1}{7\times1}\)
\(=\frac{1}{7}\)