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123456789 , ta co:
2014^9^9^123456789 & 2015^123456789^123456789^9
ta thay 2014 <2015 \ 9<1...9 \9< 1...9
nen 2014^9^9^a < 2015^a^a^9
vi may minh mat vietkey nen khong dang dau thong cam
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
A=92013+1/92014+1
9A=92014+9/92014+1
=(92014+1/92014+1)+(8/92014+1)
=1+8/92014+1
B=92014+1/92015+1
9B=92015+9/92015+1
=(92015+1/92015+1)+(8/92015+1)
=1+8/92015+1
Vì 8/92014+1 > 8/92015+1 nên A>B
**** bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 : a ) - 2015 + 2016 - ( - 2015 )
= - 2015 + 2016 +2015
= [ ( - 2015 ) + 2015 ] + 2016
= 0 + 2016
= 2016
b ) 18 . 17 - 3 . 6 . 7 = 18 . 17 - 18 . 7 = 18.( 17 - 7 ) = 18 . 10 = 180
Câu 2 : a ) 2x - 19 = 9 <=> 2x = 9 + 19 <=> 2x = 28 => x = 14
b ) 37 - 3x = 7 <=> 3x = 37 - 7 <=> 3x = 30 => x = 10
c ) Vì | x - 3 | = 2 <=> x - 3 = + 2
TH1 : x - 3 = 2 <=> x = 2 + 3 => x = 5
TH2 : x - 3 = - 2 <=> x = - 2 + 3 => x = 1
Vậy x = { 5 ; 1 }
Câu 3 : Vì 2016 = 2016 mà 2014 < 2016 => A > B
Câu 1 :
a) -2015 + 2016 - ( -2015)
= -2015 + 2016 + 2015
= [(-2015) + 2015 ] + 2016
= 2016
b) 18.17 - 3.6.7
= 18.17 - 18.7
= 18.(17-7)
= 18.10
= 180
Câu 2 :
a) 2x - 19 = 9
=> 2x = 28
=> x = 14
b) 37 - 3x = 7
=> 3x = 30
=> x = 10
c) |x-3| = 2
=> x - 3 = 2 hoặc x - 3 = -2
Với x - 3 = 2 => x = 5
Với x - 3 = -2 => x = 1
Vậy x = 5 hoặc x = 1
Câu 3 :
Nhận thấy : 2014 < 2016
=> 2016.2016 > 2014.2016
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
So sánh A và B biết :
A = a^a^9 = a9.a
B = 9^9^a = 99.a
Đây là hai lũy thừa cùng số mũ
mà a = 123456789 > 9 suy ra A > B nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
phân tích B ta có
B = \(\frac{2014+2015}{2015+2016}=\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)
vì \(\frac{2014}{2015+2016}<\frac{2014}{2015}\) ., \(\frac{2015}{2015+2016}<\frac{2015}{2016}\)
=> B< A
A=2014/2015+2015/2016. B=(2014+2015)/(2015+2016)
A=1-1/2015+1-1/2016. B=1-2/4031
A=1+1-(2015+2016)/(2015x2016). So sánh
A=1+1-(4031)/(2015x2x1008). 1+1-[4031/(4030x1008)]>1;1-2/4031<1.
A=1+1-[4031/(4030x1008)]. Vậy 1+1-[4031/(4030x1008)]>1-2/4031.
=>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có 1+2-3-4+5+6-7-8+9+10-...+2013+2014-2015-2016
= 1+ (2-3) + (-4+5) + (6-7)+(-8+9) +...+(-2013+2014)+(2015-2016)
= 1 + (-1) + 1 + (-1) + 1 +...+ 1 + -1
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2014^{2015}+2}{2014^{2016}+9}\)
\(2014A=\frac{2014\left(2014^{2015}+2\right)}{2014^{2016}+9}=\frac{2014^{2016}+4028}{2014^{2016}+9}=\frac{\left(2014^{2016}+9\right)+4019}{2014^{2016}+9}=\frac{2014^{2016}+9}{2014^{2016}+9}+\frac{4019}{2014^{2016}+9}=1+\frac{4019}{2014^{2016}+9}\)
\(B=\frac{2014^{2016}+2}{2014^{2017}+9}\)
\(2014B=\frac{2014\left(2014^{2016}+2\right)}{2014^{2017}+9}=\frac{2014^{2017}+4028}{2014^{2017}+9}=\frac{2014^{2017}+9+4019}{2014^{2017}+9}=\frac{2014^{2017}+9}{2014^{2017}+9}+\frac{4019}{2014^{2017}+9}=1+\frac{4019}{2014^{2017}+9}\)
Ta thấy:
\(2014^{2016}+9< 2014^{2017}+9\)
\(\Rightarrow\frac{4019}{2014^{2016}+9}>\frac{4019}{2014^{2017}+9}\)
\(\Rightarrow1+\frac{4019}{2014^{2016}+9}>1+\frac{4019}{2014^{2017}+9}\)
\(\Rightarrow A>B\)
Vậy ....
thay a boi 123456789 , ta co:
2014^9^9^123456789 & 2015^123456789^123456789^9
ta thay 2014 <2015 \ 9<1...9 \9< 1...9
nen 2014^9^9^a < 2015^a^a^9