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Giải:
Ta gọi \(\dfrac{10^{1990}+1}{10^{1991}+1}\) =A và \(\dfrac{10^{1991}}{10^{1992}}\) =B
Ta có:
A=\(\dfrac{10^{1990}+1}{10^{1991}+1}\)
10A=\(\dfrac{10^{1991}+10}{10^{1991}+1}\)
10A=\(\dfrac{10^{1991}+1+9}{10^{1991}+1}\)
10A=\(1+\dfrac{9}{10^{1991}+1}\)
Tương tự:
B=\(\dfrac{10^{1991}}{10^{1992}}\)
10B=\(\dfrac{10^{1992}}{10^{1992}}=1\)
Vì \(\dfrac{9}{10^{1991}+1}< 1\) nên 10A<10B
⇒ \(\dfrac{10^{1990}+1}{10^{1991}+1}\) < \(\dfrac{10^{1991}}{10^{1992}}\)
Có :
A = 10 - 9/10^1991+1
B = 10 - 9/10^1992+1
Vì 10^1991+1 < 10^1992+1 => 9/10^1991+1 > 9/10^1992+1
=> A < B
Tk mk nha
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
\(\frac{A}{10}=\frac{10^{1992}+1}{10^{1992}+10}=\frac{\left(10^{1992}+10\right)-9}{10^{1992}+10}=1-\frac{9}{10^{1992}+10}\)
\(\frac{B}{10}=\frac{10^{1993}+1}{10^{1993}+10}=\frac{\left(10^{1993}+10\right)-9}{10^{1993}+10}=1-\frac{9}{10^{1993}+10}\)
Vì \(10^{1992}+10< 10^{1993}+10\) nên \(1+\frac{9}{10^{1993}+10}>1+\frac{9}{10^{1993}+10}\)
Do đó \(A>B\)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
A= 10^1992+1/10^1991+1
10/A= 10^1992+1/10^1990+10
=1-9/10^1992+10
B=10^1993+1/10^1993+1
10/B=10^1993+1/10^1993+10
=1-9/10^1993+10
Vi 9/10^99+10>9/10^1993+10
nen 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