so sánh (1/6)^10 và (1/2)^50
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C1:A = \(\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)
= \(1+\frac{3}{10^{50}-1}\)
B = \(\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)
= \(1+\frac{3}{10^{50}-3}\)
Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)=) \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)=) \(A< B\)
C2: Áp dụng tính chất : Nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Vì B > 1 =) B > \(\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
(=) B > A
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(\(\frac{1}{2}\))50=(\(\frac{1}{2^5}\))10=(\(\frac{1}{32}\))10
Do 1/6> 1/30 nên (\(\frac{1}{6}\))10>(\(\frac{1}{2}\))50
\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left[\frac{1^5}{2^5}\right]^{10}=\left[\frac{1}{32}\right]^{10}\)
Vì 2 phân số này có cùng tử mà 6 < 30
=> \(\frac{1}{6}>\frac{1}{30}\)
=> \(\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
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1,1020và 9010
ta có:+,1020=(102)10=10010
+,9010=9010
vì 10010>9010=>1020>9010
2,(1/16)10 và (1/2)50
ta có:+, (1/16)10=(1/16)10
+,(1/2)50=(1/25)10=(1/32)10
vì (1/16)10>(1/32)10=>(1/16)10>(1/2)50
k mik nhé
\(a,\) \(10^{20}=10^{10+10}=10^{10}.10^{10}\)
\(90^{10}=9^{10}.10^{10}\)
Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)
\(\Rightarrow10^{20}>90^{10}\)
Vậy \(10^{20}>90^{10}\)
\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)
Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
~~~~~~~~~~Hok tốt~~~~~~~~~~~
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\(\left(\dfrac{1}{2}\right)^{50}=\left[\left(\dfrac{1}{2}\right)^5\right]^{10}=\left(\dfrac{1}{32}\right)^{10}\)
1/12>1/32
=>(1/12)^10>(1/32)^10
=>(1/12)^10>(1/2)^50
Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)
\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}=\dfrac{1}{\left(2^5\right)^{10}}=\dfrac{1}{32^{10}}\)
Do \(12< 32\Rightarrow12^{10}< 32^{10}\)
\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)
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Ta có:
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)
Vì \(10^{50}-1>10^{50}-3\Rightarrow\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)(2 phân số có cùng tử số, mẫu số của phân số nào lớn hơn thì phân
số đó nhỏ hơn)
\(\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}.\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}.\)
Do 1050-1 > 1050-3 ; => \(1+\frac{3}{10^{50}-3}>1+\frac{3}{10^{50}-1}\)
=> B > A
\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5.10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}=\frac{1}{32^{10}}