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Ta thấy \(\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vậy là em so sánh được rồi nhé :)
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
M=\(\frac{10^8-1+3}{10^8-1}\)=1+\(\frac{3}{10^8-1}\)
N=\(\frac{10^8-3+3}{10^8-3}\)=1+\(\frac{3}{10^8-3}\)
Ta có:\(\frac{3}{10^8-1}\)<\(\frac{3}{10^8-3}\) NÊN M<N
Đề hình như sai rùi bn, ở A mẫu phải là 108 - 1 chứ
Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
Ta có:
\(B=\frac{10^8}{10^8-3}< \frac{10^8+2}{10^8-3+2}=\frac{10^8+2}{10^8-1}=A\)
=> B < A
A =\(\dfrac{10^8+2}{10^8-1}\)= 1\(\dfrac{3}{10^8-1}\)
B=\(\dfrac{10^8}{10^8-3}\)=1\(\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}\)<\(\dfrac{3}{10^8-3}\)
nên A<B
\(n\left(n+3\right)=n^2+3n\)
\(\left(n+2\right)\left(n+1\right)=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\left(n\in N\right)\)
b) \(\dfrac{n}{2n+1}=\dfrac{3n}{6n+3}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}=1+\dfrac{1}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\left(1+\dfrac{3}{10^8-3}\right)\)
Vì \(\dfrac{1}{10^8-1}>\dfrac{3}{10^8-3}\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)
Làm dần dần và làm từ từ, suy ra được nhiều cách giải.
a) \(\dfrac{n}{n+1}\) và \(\dfrac{n+2}{n+3}\)
+ Cách 1:
\(\dfrac{n}{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac{1}{n+1}\)
\(\dfrac{n+2}{n+3}=\dfrac{n+3-1}{n+3}=1-\dfrac{1}{n+3}\)
Vì \(\dfrac{1}{n+1}>\dfrac{1}{n+3}\) nên \(1-\dfrac{n}{n+1}< 1-\dfrac{1}{n+3}\)
\(\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
+ Cách 2:
Ta so sánh: \(n\left(n+3\right)\) và \(\left(n+1\right)\left(n+2\right)\)
\(n\left(n+3\right)=nn+3n=n^2+3n\)
\(\left(n+1\right)\left(n+2\right)=\left(n+1\right)n+\left(n+1\right).2=n^2+n+2n+2=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\) nên \(\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
b) \(\dfrac{n}{2n+1}\) và \(\dfrac{3n+1}{6n+3}\)
Ta so sánh: \(n\left(6n+3\right)\) và \(\left(2n+1\right)\left(3n+1\right)\)
\(n\left(6n+3\right)=n.6n+3n=6n^2+3n\)
\(\left(2n+1\right)\left(3n+1\right)=\left(2n+1\right)3n+\left(2n+1\right)=6n^2+3n+2n+1=6n^2+5n+1\)
Vì \(6n^2+3n< 6n^2+5n+1\) nên \(\dfrac{n}{2n+1}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}\) và \(\dfrac{10^8}{10^8-3}\)
\(\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\) nên \(\dfrac{10^8+2}{10^8-1}>\dfrac{10^8}{10^8-3}\)
d) \(\dfrac{3^{17}+1}{3^{20}+1}\) và \(\dfrac{3^{20}+1}{3^{23}+1}\)
(đang tìm cách làm, và thêm vài cách khác)
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
Ta có :
\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\Rightarrow A< B\)
~ Học tốt ~
\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}< \dfrac{3}{10^8-3}\)
\(\Rightarrow1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\)
\(\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)
Vậy A < B.
\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}< \dfrac{3}{10^8-3}\)
Nên \(1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\)
Vậy A < B.
\(M=\dfrac{10^8+2}{10^8-1}=\dfrac{\left(10^8-1\right)+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(N=\dfrac{10^8}{10^8-3}=\dfrac{\left(10^8-3\right)+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(1+\dfrac{3}{10^8-3}< 1+\dfrac{3}{10^8-1}\) nên \(M< N\)