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b) \(\frac{5}{9}\)và \(\frac{5}{8}\) :Quy đồng mẫu số : \(\frac{5}{9}\) = \(\frac{5.8}{9.8}\) = \(\frac{40}{72}\) ; \(\frac{5}{8}\) = \(\frac{5.9}{8.9}\) = \(\frac{45}{72}\)
Vì \(\frac{40}{72}\) < \(\frac{45}{72}\) nên \(\frac{5}{9}\) < \(\frac{5}{8}\)
c)\(\frac{8}{7}\) và \(\frac{7}{8}\) :Quy đồng mẫu số: \(\frac{8}{7}\) = \(\frac{8.8}{7.8}\) = \(\frac{64}{56}\) ; \(\frac{7}{8}\) = \(\frac{7.7}{8.7}\) =\(\frac{49}{56}\)
Vì \(\frac{64}{56}\) > \(\frac{49}{56}\) nên \(\frac{8}{7}\) > \(\frac{7}{8}\)
bạn an đông à cái câu A của bạn sai một chút.
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a)\(\frac{3}{7}\) và\(\frac{2}{8}\) :Quy đồng mẫu số : \(\frac{3}{7}\) = \(\frac{3.8}{7.8}\) = \(\frac{24}{56}\) ; \(\frac{2}{8}\) = \(\frac{2.7}{8.7}\) = \(\frac{14}{56}\)
Vì \(\frac{24}{56}\) > \(\frac{14}{56}\) nên \(\frac{3}{7}\) > \(\frac{2}{8}\)
a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)
Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)
a) 34=3x44x4=1216,67=6x27x2=121434=3x44x4=1216,67=6x27x2=1214
Do 16 > 14 => 1216<1214hay34<67
Ta có
\(\frac{13}{27}:\frac{13}{27}=1\)
\(\frac{7}{15}:\frac{13}{27}=\frac{63}{65}\)
Mặt khác \(\frac{63}{65}< 1\)
\(\Rightarrow\frac{13}{27}:\frac{13}{27}>\frac{7}{15}:\frac{23}{27}\)
\(\Rightarrow\frac{13}{27}:\frac{13}{27}\times\frac{13}{27}>\frac{7}{15}:\frac{23}{27}\times\frac{13}{27}\)
\(\Rightarrow\frac{13}{27}>\frac{7}{15}\)
1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)
Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)
Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
Bài 1:
a: Cách 1: \(\frac37=\frac{3\cdot9}{7\cdot9}=\frac{27}{63};\frac59=\frac{5\cdot7}{9\cdot7}=\frac{35}{63}\)
mà 27<35
nên \(\frac37<\frac59\)
Cách 2: Ta có: \(\frac37=\frac{3\cdot5}{7\cdot5}=\frac{15}{35}\)
\(\frac59=\frac{5\cdot3}{9\cdot3}=\frac{15}{27}\)
mà \(\frac{15}{35}<\frac{15}{27}\left(35>27\right)\)
nên \(\frac37<\frac59\)
b: Cách 1: \(\frac{6}{12}=\frac{6\cdot3}{12\cdot3}=\frac{18}{36};\frac49=\frac{4\cdot4}{9\cdot4}=\frac{16}{36}\)
mà 18>16
nên \(\frac{6}{12}>\frac49\)
Cách 2: \(\frac{6}{12}=\frac{6\cdot2}{12\cdot2}=\frac{12}{24};\frac49=\frac{4\cdot3}{9\cdot3}=\frac{12}{27}\)
mà \(\frac{12}{24}>\frac{12}{27}\left(24<27\right)\)
nên \(\frac{6}{12}>\frac49\)
c: Cách 1: \(\frac{15}{25}=\frac{15\cdot18}{25\cdot18}=\frac{270}{450};\frac{54}{90}=\frac{54\cdot5}{90\cdot5}=\frac{270}{450}\)
Do đó: \(\frac{15}{25}=\frac{54}{90}\)
Cách 2: \(\frac{15}{25}=\frac{15:5}{25:5}=\frac35;\frac{54}{90}=\frac{54:18}{90:18}=\frac35\)
do đó: \(\frac{15}{25}=\frac{54}{90}\)
d: Cách 1: \(\frac{25}{30}=\frac{25\cdot3}{30\cdot3}=\frac{75}{90};\frac{75}{28}=\frac{75\cdot1}{28\cdot1}=\frac{75}{28}\)
mà \(\frac{75}{90}<\frac{75}{28}\left(90>28\right)\)
nên \(\frac{25}{30}<\frac{75}{28}\)
Cách 2: \(\frac{25}{30}=\frac{25\cdot14}{30\cdot14}=\frac{350}{420}\)
\(\frac{75}{28}=\frac{75\cdot15}{28\cdot15}=\frac{1125}{420}\)
mà 350<1125
nên \(\frac{25}{30}<\frac{75}{28}\)
e: cách 1: \(\frac{27}{45}=\frac{27\cdot4}{45\cdot4}=\frac{108}{180};\frac{18}{36}=\frac{18\cdot5}{36\cdot5}=\frac{90}{180}\)
mà 108>90
nên \(\frac{27}{45}>\frac{18}{36}\)
Cách 2: \(\frac{27}{45}=\frac{27\cdot2}{45\cdot2}=\frac{54}{90}\)
\(\frac{18}{36}=\frac{18\cdot3}{36\cdot3}=\frac{54}{108}\)
mà \(\frac{54}{90}>\frac{54}{108}\left(90<108\right)\)
nên \(\frac{27}{45}>\frac{18}{36}\)
\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\sqrt[3]{\left(1-\sqrt{3}\right)\left(\sqrt{3}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{3}\right)^3}\)=1-\(\sqrt{3}\)
\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}=\sqrt[3]{\left(1-\sqrt{5}\right)\left(\sqrt{5}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=1-\(\sqrt{5}\)
Ta thấy \(\sqrt{5}>\sqrt{3}\)nên 1-\(\sqrt{3}\)>\(1-\sqrt{5}\)
Vậy \(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}\)>\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
a) 1 3 = 2 6 ; 2 6 < 5 6 ⇒ 1 3 < 5 6
b) 4 5 = 28 35 ; 3 7 = 15 35 ; 28 35 > 15 35 ⇒ 4 5 > 3 7