so sanh A,B .A=\(\frac{10^9+1}{10^8+1}\) ,,B\(\frac{10^{20}+1}{10^{21}+1}\)
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25 tháng 7 2019 lúc 17:43
So sánh A và B:
a) A = \(\frac{10^{19}+1}{10^{20}+1}\); B = \(\frac{10^{20}+1}{10^{21}+1}\)
b) A = \(\frac{9^{99}+1}{9^{100}+1}\); B = \(\frac{10^{98}-1}{10^{99}-1}\)
SO SANH:
\(A=\frac{10^9+1}{10^{10}+1}\)
\(B=\frac{10^7+1}{10^8+1}\)
so sanh 2 phan so
a=10^19+1/10^20+1;b=10^20+1+10^21+1
So sánh A và B biết
A=\(\frac{10^{19}+1}{10^{20}+1}\)
B=\(\frac{10^{20}+1}{10^{21}+1}\)
10A=\(\frac{10^{20}+10}{10^{20}+1}\)=\(\frac{10^{20}+1+9}{10^{20}+1}\)=\(1\)+\(\frac{9}{10^{20}+1}\)
10B=\(\frac{10^{21}+10}{10^{21}+1}\)=\(\frac{10^{21}+1+9}{10^{21}+1}\)=\(1\)+\(\frac{9}{10^{21}+1}\)
Vì \(\frac{9}{10^{20}+1}\)>\(\frac{9}{10^{21}+1}\)nên 10A>10B\(\Rightarrow\)A>B
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So sánh A = \(\frac{10^{19}+1}{10^{20}+1}\) và B = \(\frac{10^{20}+1}{10^{21}+1}\) ?
Áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\) (a;b;c \(\in\) N*)
Ta có:
\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}\)
\(B< \frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
=> A > B
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so sanh:
A= \(\frac{20^{10}+1}{20^{10}-1}\) và B=\(\frac{20^{10}-1}{20^{10}-3}\)
SO SANH \(\frac{10^{19}+1}{10^{20}+1}\) VA \(\frac{10^{20}+1}{10^{21}+1}\)
so sánh
A=\(\frac{10^{19}+1}{10^{20}+1}\)và B=\(\frac{10^{20}+1}{10^{21}+1}\)
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
VẬY B<A
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So sanh :
A =\(\frac{20^{10}+1}{20^{10}-1}vaB=\frac{20^{10}-1}{20^{10}-3}\)
Ta có a/b >1 => a/b > a+n/b+n(a, b,n \(\in\) N*)
B = 2010-1/2010-3 > 1 nên B = 2010-1/2010-3 > 2010-1+2/2010-3+2
= 2010+1/ 2010-1 = A
Vậy B > A
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