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\(y=\frac{16.29-32}{16.37+32}=\frac{9}{13}\)
vậy y = \(\frac{9}{13}\)
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\(=\dfrac{16\cdot30-30-16}{16\cdot37+16\cdot2}=\dfrac{30\left(16-1\right)-16}{16\cdot39}\)
\(=\dfrac{30\cdot15-16}{16\cdot39}=\dfrac{434}{16\cdot39}=\dfrac{217}{312}\)
a) \(\frac{3^2.2^4}{8.3^2}=\frac{3^2.2^4}{2^3.3^2}=2\)
b)\(\frac{84.45}{49.54}=\frac{2^2.7.3^3.5}{7^2.2.3^3}=\frac{10}{7}\)
c) \(\frac{36.10.21}{2^3.63.35}=\frac{2^3.3^3.5.7}{2^3.3^2.5.7^2}=\frac{3}{7}\)
d) \(\frac{16.29-32}{16.37+32}=\frac{16.29-16.2}{16.37+2.16}=\frac{16.\left(29-2\right)}{16.\left(37+2\right)}=\frac{27}{39}=\frac{9}{13}\)
Ta có:
\(\begin{array}{l}\frac{{24}}{{108}} = \frac{{24:12}}{{108:12}} = \frac{2}{9};\\\frac{{80}}{{32}} = \frac{{80:16}}{{32:16}} = \frac{5}{2}\end{array}\)
\(\frac{48}{168}\cdot\frac{132}{32}=\frac{2^4\cdot3}{2^3\cdot3\cdot7}\cdot\frac{2^2\cdot3\cdot11}{2^5}=\frac{2\cdot1}{1\cdot1\cdot7}\cdot\frac{1\cdot3\cdot11}{2^3}=\frac{33}{28}\)
A = 1 + 3 + 32 + 33 + ... + 3100
3A = 3 + 32 + 33 +34+ .... + 3101
3A - A = (3 + 32 + 34 + ... + 3101) - (1 + 3 + 32 + 33 + ... + 3100)
2A = 3 + 32 + 34 + ... + 3101 - 1 - 3 - 32 - 33 - ... - 3100
2A = (3 - 3) + (32 - 32) + ... + (3100 - 3100) + (3101 - 1)
2A = 3101 - 1
A = \(\dfrac{3^{101}-1}{2}\)
\(\frac{16.29-32}{16.37+32}\)
\(=\frac{16.29-16.2}{16.37+16.2}\)
\(=\frac{16\left(29-2\right)}{16\left(37+2\right)}\)
\(=\frac{16.27}{16.39}\)
\(=\frac{27}{39}=\frac{9}{13}\)