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KK
5 tháng 6 2016
Xét 3 trường hợp
TH1 \(a=b\)\(\Rightarrow\frac{a}{b}=1=\frac{a+2016}{b+2016}\)
TH2 \(a>b\Rightarrow\frac{a}{b}>1\)\(\Rightarrow2016a>2016b\)
Ta có: \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016a}{b\left(b+2016\right)}\)
\(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016b}{b\left(b+2016\right)}\)
Ta có: 2016a>2016b => ab+2016a>ab+2016b hay a/b>a+2016/b+20116
TH3
\(a< b\Rightarrow\frac{a}{b}< 1\)\(\Rightarrow2016a< 2016b\)
Ta có: \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016a}{b\left(b+2016\right)}\)
\(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016b}{b\left(b+2016\right)}\)
Ta có: 2016a<2016b => ab+2016a<ab+2016b hay a/b<a+2016/b+20116
9 tháng 9 2021
\(\dfrac{a}{b}=\dfrac{a\left(b+2021\right)}{b\left(b+2021\right)}=\dfrac{ab+2021a}{b\left(b+2021\right)}\\ \dfrac{a+2021}{b+2021}=\dfrac{ab+2021b}{b\left(b+2021\right)}\)
Vì \(b>0\Rightarrow b\left(b+2021\right)>0\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2021}{b+2021}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2021}{b+2021}=1\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)
DN
1
9 tháng 6 2016
Ta có: \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016a}{b\left(b+2016\right)}\) ;
\(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016b}{b\left(b+2016\right)}\)
Với a = b thì \(\frac{a}{b}=\frac{a+2016}{b+2016}\)
Với a < b thì \(\frac{a}{b}< \frac{a+2016}{b+2016}\)
Với a > b thì \(\frac{a}{b}>\frac{a+2016}{b+2016}\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$
$=1-\frac{9}{10^{2021}-1}>1$
$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$
$=1+\frac{9}{10^{2022}+1}<1$
$\Rightarrow 10A> 1> 10B$
Suy ra $A> B$
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 12 2023
B = 2020.2021.2022
B = (2021 - 1).(2021 + 1).2021
B = (2021.2021 + 2021 - 2021 - 1).2021
B = (20212021-1).2021
B = 20213 - 2021
Vậy A > B
NQ
1
14 tháng 9 2023
`# \text {DNamNgV}`
\(A=1+2+2^2+...+2^{2021}\text{ và }B=2^{2022}\)
Ta có:
\(A=1+2+2^2+...+2^{2021}\\ \Rightarrow2A=2+2^2+2^3+...+2^{2022}\\\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow A=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\\ \Rightarrow A=2^{2022}-1\)
Vì \(2^{2022}-1< 2^{2022}\)
\(\Rightarrow A< B.\)