Dạng 4. So sánh qua phần bù (hay phần thiếu) có đáp án

b) Ta có:

+) $1-\frac{1007}{1009}=\frac{1009}{1009}-\frac{1007}{1009}=\frac{2}{1009}$

+) $1-\frac{1005}{1007}=\frac{1007}{1007}-\frac{1005}{1007}=\frac{2}{1007}$

c) Ta có:

+) $1-\frac{2021}{2023}=\frac{2023}{2023}-\frac{2021}{2023}=\frac{2}{2023}$

+) $1-\frac{2017}{2019}=\frac{2019}{2019}-\frac{2017}{2019}=\frac{2}{2019}$

c) Ta có:

+) $1-\frac{2004}{2005}=\frac{2005}{2005}-\frac{2004}{2005}=\frac{1}{2005}$

+) $1-\frac{2001}{2004}=\frac{2004}{2004}-\frac{2001}{2004}=\frac{3}{2004}$

d) Ta có:

+) $1-\frac{1775}{1777}=\frac{1777}{1777}-\frac{1775}{1777}=\frac{2}{1777}$

+) $1-\frac{1768}{1771}=\frac{1771}{1771}-\frac{1768}{1771}=\frac{3}{1771}$

Ta có:

+) $1-A=1-\frac{{10}^9+1}{{10}^{10}+1}=\frac{{10}^{10}+1}{{10}^{10}+1}-\frac{{10}^9+1}{{10}^{10}+1}=\frac{{10}^{10}-{10}^9}{{10}^{10}+1}=\frac{{10}^9.9}{{10}^{10}+1}$

+) $1-B=1-\frac{{10}^8+1}{{10}^9+1}=\frac{{10}^9+1}{{10}^9+1}-\frac{{10}^8+1}{{10}^9+1}=\frac{{10}^9-{10}^8}{{10}^9+1}=\frac{{10}^8.9}{{10}^9+1}$

+) Để so sánh $1-A$ và $1-B$, ta so sánh  $\frac{10}{{10}^{10}+1}$ và $\frac{1}{{10}^9+1}$

$$\begin{gathered} \frac{1}{{10}^9+1}=\frac{10}{{10}^{10}+10}<\frac{10}{{10}^{10}+1} \\ \Rightarrow 1-B<1-A\Rightarrow A<B \end{gathered}$$

Ta có:

+) $A<1\Rightarrow \frac{7^{99}+2}{7^{100}+2}<\frac{5}{5}\Rightarrow \frac{7^{99}+1}{7^{100}+1}<\frac{7^{99}+2+5}{7^{100}+2+5}=\frac{7^{99}+7}{7^{100}+7}=\frac{7^{98}+1}{7^{99}+1}$

+) $\frac{7^{98}+1}{7^{99}+1}<1\Rightarrow \frac{7^{98}+1}{7^{99}+1}<\frac{1}{1}\Rightarrow \frac{7^{98}+1}{7^{99}+1}<\frac{7^{98}+1+1}{7^{99}+1+1}=\frac{7^{98}+2}{7^{99}+2}$

+ Vậy $A<B.$

Ta có:

$A<1\Rightarrow \frac{{11}^{89}+1}{{11}^{90}+1}<\frac{10}{10}\Rightarrow \frac{{11}^{89}+1}{{11}^{90}+1}<\frac{{11}^{89}+1+10}{{11}^{90}+1+10}=\frac{{11}^{88}+1}{{11}^{89}+1}$

$\frac{{11}^{88}+1}{{11}^{89}+1}<1\Rightarrow \frac{{11}^{88}+1}{{11}^{89}+1}<\frac{10}{10}\Rightarrow \frac{{11}^{88}+1}{{11}^{89}+1}<\frac{{11}^{88}+1+10}{{11}^{89}+1+10}=\frac{{11}^{87}+1}{{11}^{88}+1}$

Vậy $A<\frac{{11}^{88}+1}{{11}^{89}+1}<B$

Ta có: $\frac{43}{49}=\frac{43.4}{49.4}=\frac{172}{196}$

$\frac{31}{35}=\frac{31.6}{35.6}=\frac{186}{210}$

$1-\frac{43}{49}=1-\frac{172}{196}=\frac{196}{196}-\frac{172}{196}=\frac{24}{196}$

$1-\frac{31}{35}=1-\frac{186}{210}=\frac{210}{210}-\frac{186}{210}=\frac{24}{210}$

$\frac{24}{196}>\frac{24}{210}\Rightarrow \frac{43}{49}<\frac{31}{35}$

a) Ta có:  

$\frac{7}{15}<1\Rightarrow \frac{7}{15}<\frac{2}{2}\Rightarrow \frac{7}{15}<\frac{7+2}{15+2}=\frac{9}{17}<\frac{12}{17}.$

c) Ta có:

$\frac{13}{27}<\frac{13}{26}=\frac{1}{2}$

$\frac{27}{41}>\frac{27}{54}=\frac{1}{2}$

d) Ta có:

$1-\frac{1998}{1999}=\frac{1999}{1999}-\frac{1998}{1999}=\frac{1}{1999}$

$1-\frac{1999}{2000}=\frac{2000}{2000}-\frac{1999}{2000}=\frac{1}{2000}$

e) Ta có

$\frac{23}{47}<\frac{23}{46}=\frac{1}{2}$

$\frac{24}{45}>\frac{24}{48}=\frac{1}{2}$

g) Ta có:

$\frac{13}{27}<\frac{13}{26}=\frac{1}{2}$

$\frac{17}{33}>\frac{17}{34}=\frac{1}{2}$

b) Ta có

$\frac{13}{60}<\frac{15}{60}=\frac{1}{4}$

$\frac{27}{100}>\frac{25}{100}=\frac{1}{4}$

c) Ta có:

d) Ta có:

$47>15\Rightarrow \frac{47}{15}>1$

$29<35\Rightarrow \frac{29}{35}<1$

a) Ta có: $\frac{13}{15}=\frac{13.111}{15.111}=\frac{1443}{1665}.$

$1-\frac{1443}{1665}=\frac{1665}{1665}-\frac{1443}{1665}=\frac{222}{1665}$

$1-\frac{1333}{1555}=\frac{1555}{1555}-\frac{1333}{1555}=\frac{222}{1555}$

$\frac{222}{1665}<\frac{222}{1555}\Rightarrow \frac{13}{15}>\frac{1333}{1555}$

b) Ta có: $1-\frac{42}{43}=\frac{1}{43};1-\frac{58}{59}=\frac{1}{59}$

a) Ta có: $\frac{34.34}{33.35}=\frac{{34}^2}{(34-1)(34+1)}=\frac{{34}^2}{{34}^2-1}<1.$

b) Ta có: $\frac{1999.1999}{1995.1995}=\frac{{1999}^2}{{1995}^2}>1$

c) Ta có: $\frac{198519851985.198719871987}{198619861986.198619861986}=\frac{\mathrm{1985.100010001.1987.100010001}}{\mathrm{1986.100010001.1986.100010001}}=\frac{1985.1987}{{1986}^2}$

$\begin{array}{l}=\frac{(1986-1)(1986+1)}{{1986}^2}\\ =\frac{{1986}^2-1}{{1986}^2}<1\end{array}$

a) Ta có:    1 - $\frac{2012}{2013}$ = $\frac{1}{2013}$;                    

                    1 -  $\frac{2013}{2014}$= $\frac{1}{2014}$

Vì  $\frac{1}{2013}>\frac{1}{2014}$ nên  $\frac{2012}{2013}<\frac{2013}{2014}$ .

b)  Ta thấy: $\frac{1006}{1007}=\frac{1006\times 2}{1007\times 2}=\frac{2012}{2014}$

Ta có: $1-\frac{2012}{2014}=\frac{2}{2014}$;    $1-\frac{2013}{2015}=\frac{2}{2015}$

Vì $\frac{2}{2014}>\frac{2}{2015}$ nên $\frac{2012}{2014}<\frac{2013}{2015}$ hay $\frac{1006}{1007}<\frac{2013}{2015}$

c) Ta thấy:  $\frac{64}{73}=\frac{64\times 2}{73\times 2}=\frac{128}{146}$ ;   $\frac{45}{51}=\frac{45\times 3}{51\times 3}=\frac{135}{153}$.

Ta có: $1-\frac{128}{146}=\frac{18}{146}\text{;}$   $1-\frac{135}{153}=\frac{18}{153}.$            .

d) Ta thấy: $\frac{2323}{2424}=\frac{23\times 101}{24\times 101}=\frac{23}{24}$  ;       

     $\frac{20132013}{20142014}=\frac{2013\times 10001}{2014\times 10001}=\frac{2013}{2014}$

Ta có: $1-\frac{23}{24}=\frac{1}{24}\text{;}$

$1-\frac{2013}{2014}=\frac{1}{2014}.$

Vì $\frac{1}{24}>\frac{1}{2014}$nên  $\frac{23}{24}<\frac{2013}{2014}$

Vậy $\frac{2323}{2424}<\frac{20132013}{20142014}.$