So sánh:
a)A=\(\frac{10^{15}+1}{10^{16}+1}\),B=\(\frac{10^{16}+1}{10^{17}+1}\)
b)C=\(\frac{10^{1992}+1}{10^{1991}+1}\),D=\(\frac{10^{1993}+1}{10^{1992}+1}\)
K
Khách
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22 tháng 10 2017
Ta có :
\(A=\frac{10^{1992}+1}{10^{1991}+1}\)
\(\Rightarrow\frac{1}{10}A=\frac{10^{1992}+1}{10^{1992}+10}=\frac{10^{1992}+10-11}{10^{1992}+10}=1-\frac{11}{10^{1992}+10}\)
\(B=\frac{10^{1993}+1}{10^{1992}+1}\)
\(\Rightarrow\frac{1}{10}B=\frac{10^{1993}+1}{10^{1993}+10}=\frac{10^{1993}+10-11}{10^{1993}+10}=1-\frac{11}{10^{1993}+10}\)
Mà \(10^{1993}+10>10^{1992}+10\)
\(\Rightarrow\frac{11}{10^{1993}+10}< \frac{11}{10^{1992}+10}\)
\(\Rightarrow1-\frac{11}{10^{1993}+10}>1-\frac{11}{10^{1992}+10}\)
\(\Leftrightarrow\frac{1}{10}B>\frac{1}{10}A\)
\(\Rightarrow B>A\)
11 tháng 3 2018
Ta có công thức :
\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
\(\Rightarrow\)\(B>A\) hay \(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
27 tháng 11 2016
\(\Rightarrow\frac{A}{10}=\frac{10^{1992}+1}{10^{1992}+10}=\frac{10^{1992}+10-9}{10^{1992}+10}=1-\frac{9}{10\left(10^{1991}+1\right)}\)
\(\Rightarrow\frac{B}{10}=\frac{10^{1993}+1}{10^{1993}+10}=\frac{10^{1993}+10-9}{10^{1993}+10}=1-\frac{9}{10\left(10^{1992}+1\right)}\)
Vì \(1-\frac{9}{10\left(10^{1991}+1\right)}< 1-\frac{9}{10\left(10^{1992}+1\right)}\Rightarrow A< B\)
27 tháng 6 2016
\(10A=\frac{10^{1993}+10}{10^{1993}+1}=1+\frac{9}{10^{1993}+1}\)
\(10B=\frac{10^{1994}+10}{10^{1994}+1}=1+\frac{9}{10^{1994}+1}\)
\(10^{1993}+1< 10^{1994}+1\Rightarrow\frac{9}{10^{1993}+1}>\frac{9}{10^{1994}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
12 tháng 8 2017
Ta có B=\(\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}\)
= \(\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
=> B > A
22 tháng 7 2018
Ta có: \(\frac{n}{n+1}< 1\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)
\(\Rightarrow A< B\)
b. mình ko biết làm
c. mình cũng ko biết làm
d.Ta có :\(\frac{10^{1993}+1}{10^{1992}+1}>1\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}.10+10.1}{10^{1991}.10+10.1}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}+1}{10^{1991}+1}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt nhé
TN
1
DT
Đỗ Thanh Hải
CTVVIP
12 tháng 8 2017
Áp dụng tính chất :
\(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có :
\(B=\dfrac{10^{1993}+1}{10^{1992}+1}>\dfrac{10^{1993}+1+9}{10^{1992}+1+9}=\dfrac{10^{1993}+10}{10^{1992}+10}=\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=A\)
\(\Leftrightarrow B>A\)
a) Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
\(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(\frac{1}{10}C=\frac{10^{1992}+1}{10^{1992}+10}=1+\frac{10^{1992}+1}{9}\)
\(\frac{1}{10}D=\frac{10^{1993}+1}{10^{1993}+10}=1+\frac{10^{1993}+1}{9}\)
\(\frac{10^{1992}+1}{9}< \frac{10^{1993}+1}{9}\Rightarrow1+\frac{10^{1992}+1}{9}< 1+\frac{10^{1993}+1}{9}\)
\(\Rightarrow\frac{1}{10}C< \frac{1}{10}D\)
\(\Rightarrow C< D\)
Vậy C < D