so sánh p/s :
a) 7 phần 8 với 212 phần 243
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TL
a)\(\frac{8}{7};\frac{7}{8}\)
Cách 1:\(\frac{8}{7}\) và \(\frac{7}{8}\) (MSC:56)
\(\frac{8}{7}=\frac{8x8}{7x8}=\frac{64}{56}\); \(\frac{7}{8}=\frac{7x7}{8x7}=\frac{49}{56}\)
Vì \(\frac{64}{56}>\frac{49}{56}\) nên \(\frac{8}{7}>\frac{7}{8}\)
Cách 2: Vì \(\frac{8}{7}>1;\frac{7}{8}< 1\) nên \(\frac{8}{7}>\frac{7}{8}\)
b)
Cách 1: \(\frac{9}{5};\frac{5}{8}\)(MSC:40)
\(\frac{9}{5}=\frac{9x8}{5x8}=\frac{72}{40};\frac{5}{8}=\frac{5x5}{8x5}=\frac{25}{40}\)
Vì \(\frac{72}{40}>\frac{25}{40}\)nên \(\frac{9}{5}>\frac{5}{8}\)
Cách 2: Vì \(\frac{9}{5}>1;\frac{5}{8}< 1\) nên \(\frac{9}{5}>\frac{5}{8}\)
HT
a) Ta có :
\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)
\(\Rightarrow27^{27}>243^{13}\)
\(\Rightarrow-27^{27}< -243^{13}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)
b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)
c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)
\(\Rightarrow4^{50}>8^{30}\)
d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)
\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)
a) Ta có :
2727>2726=(272)13=72913>243132727>2726=(272)13=72913>24313
⇒2727>24313⇒2727>24313
⇒−2727<−24313⇒−2727<−24313
⇒(−27)27<(−243)13⇒(−27)27<(−243)13
b) (18)25>(18)26=(182)13=(164)13>(1128)13(81)25>(81)26=(821)13=(641)13>(1281)13
⇒(18)25>(1128)13⇒(81)25>(1281)13
⇒(−18)25<(−1128)13⇒(−81)25<(−1281)13
c) 450=(45)10=102410450=(45)10=102410
830=(83)10=51210<102410830=(83)10=51210<102410
⇒450>830⇒450>830
d) (19)17<(19)12<(127)12(91)17<(91)12<(271)12
⇒(19)17<(127)12⇒(91)17<(271)12
Ta có :
\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)
\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)
mình chắc chắn luôn
ta thấy \(\frac{1}{20}\)<\(\frac{1}{3}\)
thì \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{20}\)+...+\(\frac{1}{20}\)<\(\frac{1}{3}\)
vậy \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{3}\)
7/8 > 212/243
Ta có: \(\frac{7}{8}=\frac{1701}{1944}và\frac{212}{243}=\frac{1696}{1944}\)
Vì \(\frac{1701}{1944}>\frac{1696}{1944}nên\frac{7}{8}>\frac{212}{243}\)
Vậy \(\frac{7}{8}>\frac{212}{243}\)