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Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
=> \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
=> \(B< \frac{10^{1991}+10}{10^{1992}+10}\)
=> \(B< \frac{10.\left(10^{1990}+1\right)}{10.\left(10^{1991}+1\right)}\)
=> \(B< \frac{10^{1990}+1}{10^{1991}+1}=A\)
=> B < A
Ta có :
\(10A=\dfrac{10\left(10^{1990}+1\right)}{10^{1991}+1}=\dfrac{10^{1991}+10}{10^{1991}+1}=\dfrac{10^{1991}+1+9}{10^{1991}+1}=1+\dfrac{9}{10^{1991}+1}\left(1\right)\)
\(10B=\dfrac{10\left(10^{1991}+1\right)}{10^{1992}+1}=\dfrac{10^{1992}+10}{10^{1992}+1}=\dfrac{10^{1992}+1+9}{10^{1992}+1}=1+\dfrac{9}{10^{1992}+1}\left(2\right)\)
Lại có : \(1+\dfrac{9}{10^{1991}+1}>1+\dfrac{9}{10^{1992}+1}\)
\(\Leftrightarrow10A>10B\Leftrightarrow A>B\)
Vậy...
Ta có : \(A=\frac{10^{1990}+1}{10^{1991}+1}=>10A=\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=>10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{\left(10^{1991}+1\right)+9}{10^{1991}+1}\)
\(=>10A=1+\frac{9}{10^{1991}+1}\)
Ta lại có : \(B=\frac{10^{1991}+1}{10^{1992}+1}=>10B=\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
Tương tự như A => \(10B=1+\frac{9}{10^{1992}+1}\)
Vì \(\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}=>10A>10B\)
\(=>A>B\)
Ta có :
A = \(\frac{10^{1990}+1}{10^{1991}+1}\)
10A = \(\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+10}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+1+9}{10^{1991}+1}\)
10A = \(1+\frac{9}{10^{1991}+1}\left(1\right)\)
Ta lại có :
B = \(\frac{10^{1991}+1}{10^{1992}+1}\)
10B = \(\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+10}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+1+9}{10^{1992}+1}\)
10B = \(1+\frac{9}{10^{1992}+1}\left(2\right)\)
Từ \(\left(1\right)va\left(2\right)\)
Ta có :\(1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\)10A > 10B
\(\Rightarrow\)A > B
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Vì \(10^{1991}< 10^{1992}\Rightarrow1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\frac{10^{1990}+1}{10^{1991}+1}>\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow A>B\)
Ta có : \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Mà : \(\frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}\)
\(=\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)
\(=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow B< A\)
Đặt \(A=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow10A=\frac{10\cdot(10^{1990}+1)}{10^{1991}+1}\)
\(=\frac{10^{1991}+10}{10^{1991}+1}=\frac{10^{1991}+1+9}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
Đặt \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow10B=\frac{10\cdot(10^{1991}+1)}{10^{1992}+1}=\frac{10^{1992}+10}{10^{1992}+1}=\frac{10^{1992}+1+9}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Tự so sánh được rồi -_-
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(A=\dfrac{10^{1992}+1}{10^{1993}+1}< \dfrac{10^{1992}+1+9}{10^{1993}+1+9}=\dfrac{10^{1992}+10}{10^{1993}+10}\)
\(=\dfrac{10\left(10^{1991}+1\right)}{10\left(10^{1992}+1\right)}=\dfrac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow\dfrac{10^{1992}+1}{10^{1993}+1}< \dfrac{10^{1991}+1}{10^{1992}+1}\)
Hay \(A>B\)
Ta đi so sánh:
\(\dfrac{1}{A}=\dfrac{10^{1991}+1}{10^{1992}+1}\) và \(\dfrac{1}{B}=\dfrac{10^{1992}+1}{10^{1993}+1}\)
Ta có:\(\dfrac{10}{A}=\dfrac{10^{1992}+10}{10^{1992}+1}\)
\(=\dfrac{10^{1992}+1+9}{10^{1992}+1}\)
\(=1+\dfrac{9}{10^{1992}+1}\)
\(\dfrac{10}{B}=\dfrac{10^{1993}+10}{10^{1993}+1}\)
\(=\dfrac{10^{1993}+1+9}{10^{1993}+11}\)
Mà \(\dfrac{9}{10^{1992}+1}>\dfrac{9}{10^{1993}+1}\)
\(\Rightarrow\dfrac{10}{A}>\dfrac{10}{B}\)
Vậy A<B
Ta có :
\(10A=\dfrac{10^{1991}+10}{10^{1991}+1}=\dfrac{10^{1991}+1+9}{10^{1991}+1}=1+\dfrac{9}{10^{1991}+1}\)\(\left(1\right)\)
\(10B=\dfrac{10^{1992}+10}{10^{1992}+1}=\dfrac{10^{1992}+1+9}{10^{1992}+1}=1+\dfrac{9}{10^{1992}+1}\)\(\left(2\right)\)
Vì \(1+\dfrac{9}{10^{1991}+1}>1+\dfrac{9}{10^{1992}+1}\)\(\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
~ Chúc bn học tốt ~
Ta có:
A=101990+1101991+1=101990.10101991.10=101990101991=1/10A=101990+1101991+1=101990.10101991.10=101990101991=1/10 (%)
B=101991+1101992+1=101991.10101992.10=101991101992=1/10B=101991+1101992+1=101991.10101992.10=101991101992=1/10 (%) (%)