So sánh 33^9 và 11^21
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a) Ta có: \(8^{28}=2^{84}=16^{21}\)
Mà \(16>15\Rightarrow16^{21}>15^{21}\Rightarrow8^{28}>15^{21}\)
Vậy...
b) \(5^{91}>5^{90}=125^{30}\) \(\left(1\right)\)
\(11^{59}< 11^{60}=121^{30}\) \(\left(2\right)\)
Lại có: \(125>121\) \(\left(3\right)\)
Từ (1), (2) và (3) suy ra \(5^{91}>11^{59}\)
a. \(\dfrac{10}{21}>1\)
\(\dfrac{9}{23}< 1\)
\(\Rightarrow\dfrac{10}{21}>\dfrac{9}{23}\)
b. \(\dfrac{32}{33}>\dfrac{31}{33}>\dfrac{31}{34}\Rightarrow\dfrac{32}{33}>\dfrac{31}{34}\)
c. \(\dfrac{44}{47}< \dfrac{45}{47}< \dfrac{45}{46}\Rightarrow\dfrac{44}{47}< \dfrac{45}{46}\)
\(d.\dfrac{70}{117}< \dfrac{70}{115}=\dfrac{14}{23}\Rightarrow\dfrac{70}{117}< \dfrac{14}{23}< \dfrac{15}{23}\)
\(\Rightarrow\dfrac{70}{117}< \dfrac{15}{23}\)
a)1714>1614=256>3211=222>3111
b)102330<102430=2300<2305=3261<3361
c)8217>8117=368>363=2721>2621
a:Cách 1: \(\frac{21}{35}=\frac{21:7}{35:7}=\frac35\)
\(\frac{12}{40}=\frac{12:4}{40:4}=\frac{3}{10}\)
mà \(\frac35>\frac{3}{10}\left(5<10\right)\)
nên \(\frac{21}{35}>\frac{12}{40}\)
Cách 2: \(\frac{21}{35}=\frac{21\times8}{35\times8}=\frac{168}{280}\)
\(\frac{12}{40}=\frac{12\times7}{40\times7}=\frac{84}{280}\)
mà 168>84
nên \(\frac{21}{35}>\frac{12}{40}\)
b: Cách 1: \(\frac{35}{49}=\frac{35:7}{49:7}=\frac57\)
\(\frac{40}{64}=\frac{40:8}{64:8}=\frac58\)
mà \(\frac57>\frac58\left(7<8\right)\)
nên \(\frac{35}{49}>\frac{40}{64}\)
Cách 2: \(\frac{35}{49}=\frac{35\times64}{49\times64}=\frac{2240}{3136}\)
\(\frac{40}{64}=\frac{40\times49}{64\times49}=\frac{1960}{3136}\)
mà 2240>1960
nên \(\frac{35}{49}>\frac{40}{64}\)
c: Cách 1: \(\frac{9}{11}=1-\frac{2}{11}\)
\(\frac{13}{15}=1-\frac{2}{15}\)
Ta có: 11<15
=>\(\frac{2}{11}>\frac{2}{15}\)
=>\(-\frac{2}{11}<-\frac{2}{15}\)
=>\(-\frac{2}{11}+1<-\frac{2}{15}+1\)
=>\(\frac{9}{11}<\frac{13}{15}\)
Cách 2: Ta có: \(\frac{9}{11}=\frac{9\times15}{11\times15}=\frac{135}{165}\)
\(\frac{13}{15}=\frac{13\times11}{15\times11}=\frac{143}{165}\)
mà 135<143
nên \(\frac{9}{11}<\frac{13}{15}\)
d: Cách 1: \(\frac{21}{7}=\frac{21\times5}{7\times5}=\frac{105}{35}\)
mà 105>24
nên \(\frac{21}{7}>\frac{24}{35}\)
cách 2: Ta có: \(\frac{21}{7}=3>1;1>\frac{24}{35}\)
Do đó: \(\frac{21}{7}>\frac{24}{35}\)
e: Cách 1: \(\frac12=\frac{1\times2424}{2\times2424}=\frac{2424}{4848}\)
Cách 2: \(\frac{2424}{4848}=\frac{2424:2424}{4848:2424}=\frac12\)
f: Vì 19>11
nên \(\frac{19}{15}>\frac{11}{15}\)
\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)
\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)
\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)
\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)
Dễ thấy \(A=\frac{147}{10}>1\)
Mà \(B=\frac{43}{48}< 1\)
=> tự so sánh


33^9<11^21
339 > 1121