Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 $B>A$
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.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
Giải:
a)Ta có:
C=1957/2007=1957+50-50/2007
=2007-50/2007
=2007/2007-50/2007
=1-50/2007
D=1935/1985=1935+50-50/1985
=1985-50/1985
=1985/1985-50/1985
=1-50/1985
Vì 50/2007<50/1985 nên -50/2007>-50/1985
⇒C>D
b)Ta có:
A=20162016+2/20162016-1
A=20162016-1+3/20162016-1
A=20162016-1/20162016-1+3/20162016-1
A=1+3/20162016-1
Tương tự: B=20162016/20162016-3
B=1+3/20162016-3
Vì 20162016-1>20162016-3 nên 3/20162016-1<3/20162016-3
⇒A<B
Chúc bạn học tốt!
Làm tiếp:
c)Ta có:
M=102018+1/102019+1
10M=10.(102018+1)/202019+1
10M=102019+10/102019+1
10M=102019+1+9/102019+1
10M=102019+1/102019+1 + 9/102019+1
10M=1+9/102019+1
Tương tự:
N=102019+1/102020+1
10N=1+9/102020+1
Vì 9/102019+1>9/102020+1 nên 10M>10N
⇒M>N
Chúc bạn học tốt!
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a)\(12< 13;49>47\)
\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)
b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)
c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)
d)
\(1-\dfrac{67}{77}=\dfrac{10}{77}\)
\(1-\dfrac{73}{83}=\dfrac{10}{83}\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)
\(1-\dfrac{123}{128}=\dfrac{5}{128}\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)
b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)
\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)
c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)
d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)
\(\dfrac{73}{83}+\dfrac{10}{83}=1\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)
\(\dfrac{123}{128}+\dfrac{5}{128}=1\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
A = \(\dfrac{n^9+1}{n^{10}+1}\)
\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n - \(\dfrac{n-1}{n^9+1}\)
B = \(\dfrac{n^8+1}{n^9+1}\)
\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) = n - \(\dfrac{n-1}{n^8+1}\)
Vì n > 1 ⇒ n - 1> 0
\(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)
⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)
⇒ A < 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:
$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$
$\Rightarrow A< B$
choáng
dài quá mik ko làm âu