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a) Ta có : \(\frac{a}{b}=\frac{a\left(b+c\right)}{b\left(b+c\right)}=\frac{ab+ac}{b\left(b+c\right)}\)
\(\frac{a+c}{b+c}=\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+bc}{b\left(b+c\right)}\)
Vì 0<a<b nên ab+ac<ab+bc
\(\Rightarrow\frac{ab+ac}{b\left(b+c\right)}>\frac{ab+bc}{b\left(b+c\right)}\)
hay \(\frac{a}{b}< \frac{a+c}{b+c}\)
Vậy \(\frac{a}{b}< \frac{a+c}{b+c}\)
a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A
Vậy A > B
b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A
Vậy A < B.
NHỚ K CHO MK VỚI NHÉ !!!!!!!!
Lời giải:
a) Sửa đề thành CM \(\frac{a}{b}< \frac{a+m}{b+m}\) mới đúng nhé.
\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{ab+am-b(a+m)}{b(b+m)}=\frac{m(a-b)}{b(m+b)}\)
Do $\frac{a}{b}< 1; b>0\Rightarrow a< b\Rightarrow a-b< 0$
$m>0; b>0\Rightarrow \frac{m}{b(m+b)}>0$
$\Rightarrow \frac{a}{b}-\frac{a+m}{b+m}=\frac{m(a-b)}{b(m+b)}< 0$
$\Rightarrow \frac{a}{b}< \frac{a+m}{b+m}$ (đpcm)
b)
\(0< B=\frac{2020^{2017+1}}{2020^{2018+1}}=\frac{2020^{2018}}{2020^{2019}}< 1\) do $0< 2020^{2018}< 2020^{2019}$
$1>0$
$\Rightarrow B< \frac{2020^{2018}+1}{2020^{2019}+1}$ (theo phần a)
Hay $B< A$
\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)
\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)
Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)
=> 10A>10B
=>A>B
a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
Có 3 trường hợp :
* Nếu \(a>b\)
\(\Leftrightarrow\)\(ab=ab\)
\(\Leftrightarrow\)\(ab+a>ab+b\)\(a>b\)
\(\Leftrightarrow\)\(a\left(b+2018\right)>b\left(a+2018\right)\)
\(\Leftrightarrow\)\(\frac{a}{b}>\frac{a+2018}{b+2018}\)
* Nếu \(a< b\)
\(\Leftrightarrow\)\(ab=ab\)
\(\Leftrightarrow\)\(ab+b>ab+a\)\(b>a\)
\(\Leftrightarrow\)\(b\left(a+2018\right)>a\left(b+2018\right)\)
\(\Leftrightarrow\)\(\frac{a+2018}{b+2018}>\frac{a}{b}\)
\(\Leftrightarrow\)\(\frac{a}{b}< \frac{a+2018}{b+2018}\)
* Nếu \(a=b\)
\(\Rightarrow\)\(\frac{a}{b}=\frac{a}{a}=1\) \(\left(1\right)\)
\(\Rightarrow\)\(\frac{a+2018}{b+2018}=\frac{a+2018}{a+2018}=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(\frac{a}{b}=\frac{a+2018}{b+2018}\) \(\left(=1\right)\)
Vậy :
+) Nếu \(a>b\) thì \(\frac{a}{b}>\frac{a+2018}{b+2018}\)
+) Nếu \(a< b\) thì \(\frac{a}{b}< \frac{a+2018}{b+2018}\)
+) Nếu \(a=b\) thì \(\frac{a}{b}=\frac{a+2018}{b+2018}\)
Chúc bạn học tốt ~
Mk đg cần gấp.Các bn giúp mk nha.Cảm ơn các bn ^.^
Cảm ơn bn Phùng Minh Quân nha ^,^