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Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
Lời giải:
a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)
Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$
Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$
Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$
b) Rõ ràng $10^{11}-1< 10^{12}-1$.
Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$
Áp dụng kết quả phần a:
$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$
a)\(\dfrac{19}{10}>\dfrac{10}{11}\)
b)\(\dfrac{11}{10}=\dfrac{12}{11}\)
c)\(\dfrac{9}{10}< \dfrac{10}{11}\)
a) \(< \)
b) \(>\)
c) \(< \)
d) \(>\)
e) \(< \)
g) \(>\)
h) \(>\)
k) \(>\)
Lời giải:
$B=\frac{10^{11}+10}{10^{12}+10}$
Đặt $10^{11}-1=a; 10^{12}-1=b$ thì $0< a< b$. Khi đó:
$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$
$\Rightarrow A< B$
Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
a. Xét phân số trung gian là \(\dfrac{72}{78}\) , ta thấy:
\(\dfrac{72}{73}>\dfrac{72}{78}\)
\(\dfrac{58}{78}< \dfrac{72}{78}\)
\(\Rightarrow\dfrac{72}{73}>\dfrac{58}{78}\)
b. Xét phân số trung gian là \(\dfrac{n}{n+2}\) , ta thấy:
\(\dfrac{n}{n+3}< \dfrac{n}{n+2}\)
\(\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\)
\(\Rightarrow\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
c. Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< 1\) (vì tử < mẫu)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{\left(10^{11}-1\right)+11}{\left(10^{12}-1\right)+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
d. Xét phân số trung gian là \(\dfrac{1}{4}\) , ta thấy:
\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}\)
\(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\)