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\(\frac{22}{7}\)> \(\frac{11}{5}\)vì 22 : 7 = 3,14 ; 11: 5 = 2,2
\(\frac{15}{59}\)< \(\frac{24}{97}\)vì 15 : 59 = 0,21 ; 24 : 97 = 0,24
\(\frac{11}{19}\)< \(\frac{13}{18}\)vì 11 : 19 = 0,57 ; 13 : 18 = 0,72
\(\frac{7}{10}\)> \(\frac{4}{9}\)vì 7 : 10 = 0,7 ; 4 : 9 = 0,44
\(\frac{5}{9}\) . \(\frac{a}{b}\) - \(\frac{11}{21}\) = \(\frac{4}{21}\)
\(\frac{5}{9}\) . \(\frac{a}{b}\) = \(\frac{4}{21}\) + \(\frac{11}{21}\)
\(\frac{5}{9}\) . \(\frac{a}{b}\) = \(\frac{5}{7}\)
\(\frac{a}{b}\) = \(\frac{5}{7}\) : \(\frac{5}{9}\)
\(\frac{a}{b}\) = \(\frac{9}{7}\)
Vậy \(\frac{a}{b}\) = \(\frac{9}{7}\)
5/9 x a/b-11/21=4/21
<=> 5/9 x a/b=4/21+11/21
<=> 5/9 x a/b = 21/21=1
<=> a/b = 1 : 5/9
<=> a/b=9/5
Vậy a/b=9/5
a) MC: 20
\(\frac{3}{4}\)=\(\frac{3x5}{4x5}\)=\(\frac{15}{20}\)
\(\frac{2}{5}\)=\(\frac{2x4}{5x4}\)=\(\frac{8}{20}\)
b) MC: 36
\(\frac{5}{12}\)=\(\frac{5x3}{12x3}\)=\(\frac{15}{36}\)
\(\frac{11}{36}\)giữ nguyên
c) MC: 60
\(\frac{2}{3}\)=\(\frac{2x20}{3x20}\)=\(\frac{40}{60}\)
\(\frac{3}{4}\)=\(\frac{3x15}{4x15}\)=\(\frac{45}{60}\)
\(\frac{4}{5}\)=\(\frac{4x12}{5x12}\)=\(\frac{48}{60}\)
a ) Ta có :
\(\frac{3}{4}=\frac{3.5}{4.5}=\frac{15}{20}\)
\(\frac{2}{5}=\frac{2.4}{5.4}=\frac{8}{20}\)
a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b) \(\frac{99}{100}\)< \(\frac{100}{99}\); c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\);
d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)
e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)
tl :
\(\frac{5}{14}\)= \(\frac{5x21}{14x21}\)= \(\frac{105}{294}\)
\(\frac{4}{21}\)= \(\frac{4x14}{21x14}\)= \(\frac{56}{294}\)
a) \(11^{21}>9^{21}=\left(3^2\right)^{21}=3^{2.21}=3^{42}>3^{39}\)
b) \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Ta có: \(125>121\Rightarrow125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
c) \(21^{15}=\left(3.7\right)^{15}=3^{15}.7^{15}\)
\(27^5.49^8=\left(3^3\right)^5.\left(7^2\right)^8=3^{15}.7^{16}\)
Ta có: \(7^{15}< 7^{16}\Rightarrow3^{15}.7^{15}< 3^{15}.7^{16}\Rightarrow2^{15}< 27^5.49^8\)
d) \(3^{99}=\left(3^3\right)^{33}=27^{33}>11^{21}\)