so sánh
a)\(\dfrac{72}{73}\)và\(\dfrac{58}{78}\)
b)\(\dfrac{n}{n+3}\)và\(\dfrac{n+1}{n+2}\)
c)\(\dfrac{10^{11}-1}{10^{12}-1}\)và \(\dfrac{10^{10}+1}{11^{11}+1}\)
d)\(\dfrac{12}{47}\)và\(\dfrac{19}{77}\)
So sánh:
A) \(\dfrac{n+1}{n+2}\) và \(\dfrac{n}{n+3}\)
B) A= \(\dfrac{10^{11}-1}{10^{12}-1}\) và B= \(\dfrac{10^{10}+1}{10^{11}+1}\)
Mọi người giúp mình với mình đang cần gấp!
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.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
a, Cho a,b,n ϵ N* . Hãy so sánh \(\dfrac{a+n}{b+n}và\dfrac{a}{b}\)
b, Cho A= \(\dfrac{10^{11}-1}{10^{12}-1};B=\dfrac{10^{10}+1}{10^{11}+1}.\) So sánh A và 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 $B>A$
Không quy đồng ,hãy so sánh hai phân số
a \(\dfrac{19}{10}và\dfrac{10}{11}\)
b \(\dfrac{11}{10}và\dfrac{12}{11}\)
c \(\dfrac{9}{10}và\dfrac{10}{11}\)
a. 19/10 > 10/11
b. 11/10 = 12/11
c. 9/10 = 10/11
a)\(\dfrac{19}{10}>\dfrac{10}{11}\)
b)\(\dfrac{11}{10}=\dfrac{12}{11}\)
c)\(\dfrac{9}{10}< \dfrac{10}{11}\)
So sánh hai phân số:
a) \(\dfrac{1}{5}\) và \(\dfrac{3}{5}\) b) \(\dfrac{9}{10}\) và \(\dfrac{3}{10}\) c) \(\dfrac{7}{12}\) và \(\dfrac{11}{12}\) d) \(\dfrac{7}{8}\) và \(\dfrac{5}{8}\)
e) \(\dfrac{17}{100}\) và \(\dfrac{23}{100}\) g) \(\dfrac{4}{10}\) và \(\dfrac{1}{10}\) h) \(\dfrac{100}{100}\) và \(\dfrac{49}{100}\) k) \(\dfrac{15}{15}\) và \(\dfrac{2}{15}\)
a) \(< \)
b) \(>\)
c) \(< \)
d) \(>\)
e) \(< \)
g) \(>\)
h) \(>\)
k) \(>\)
So Sánh:
A=\(\dfrac{10^{11}-1}{10^{12}-1}\) và B=\(\dfrac{10^{10}+1}{10^{11}+1}\)
C=\(\dfrac{2005^{2005}+1}{2005^{2006}+1}\) và D=\(\dfrac{2005^{2004}+1}{2005^{2005}+1}\)
Cho \(A=\dfrac{10^{11}-1}{10^{12}-1}\); \(B=\dfrac{10^{10}+1}{10^{11}+1}\) So sánh \(A\) và \(B\)
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:
$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$
$\Rightarrow A< B$
So Sánh : \(\dfrac{10^{11}-1}{10^{12}-1}\)và\(\dfrac{10^{10}+1}{10^{11}+1}\)
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...
Vì \(10^{11}-1< 10^{12}-1\)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
Tính hợp lý nếu có thể:
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
c) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}\)
d) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}\)
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) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
a,\(=\dfrac{2}{9}+\left(\dfrac{-3}{10}+\dfrac{-7}{10}\right)=\dfrac{2}{9}+\left(-1\right)=\dfrac{-7}{9}\)
b,\(=\left(\dfrac{-11}{6}+\dfrac{-1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-8}{5}\)
c,\(=\dfrac{27}{13}-\left(\dfrac{106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
So sánh các phân số bằng cách chọn phân số chung gian:
a, \(\dfrac{11}{49}\) và \(\dfrac{13}{46}\)
b, \(\dfrac{62}{85}\) và \(\dfrac{73}{80}\)
c, \(\dfrac{n}{n+3}\) và \(\dfrac{n+1}{n+2}\) ( n ϵ N* )
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)