1=5
2=10
3=15
4=20
5=?
1=5
2=10
3=15
4=20
5=?
Tìm x thuộc Z, biết:
a) 5 6 + 1 6 ≤ x ≤ 13 4 + 14 8
b) − 5 6 + 8 3 + 29 − 6 ≤ x ≤ − 1 2 + 2 + 5 2
c) 79 15 + 7 5 + − 8 3 ≤ x ≤ 10 3 + 15 4 + 23 12
a) 5 6 + 1 6 ≤ x ≤ 13 4 + 14 8 ⇔ 1 ≤ x ≤ 5 ⇔ x ∈ 1 ; 2 ; 3 ; 4 ; 5
b) − 5 6 + 8 3 + 29 − 6 ≤ x ≤ − 1 2 + 2 + 5 2 ⇔ − 3 ≤ x ≤ 4 ⇔ x ∈ − 3 ; − 2 ; − 1 ; 0 ; 1 ; 2 ; 3 ; 4
c) 79 15 + 7 5 + − 8 3 ≤ x ≤ 10 3 + 15 4 + 23 12 ⇔ 4 ≤ x ≤ 9 ⇔ x ∈ 4 ; 5 ; 6 ; 7 ; 8 ; 9
x+205=102+103
So sánh: 205/103 và 74/295
Ta có: \(\dfrac{205}{103}>1;\dfrac{74}{295}< 1\Rightarrow\dfrac{205}{103}>\dfrac{74}{295}\)
1+4+7+10+...........+52+53-154
1 + 4 + 7 + 10 + ............ + 52 + 53 - 154 = 427
Đặt A=1+4+7+10+..........+52+53-154
Ta có: A=(1+4+7+............+52)+53-154
B
Ta có: B= (52+1)x18:2
=> B=477
Vậy A=477+53-154
=> A=530-154
=> A=376
Vậy tổng trên bằng 376
(2006 x (154 - 52)) - 4012
[ 2006 x ( 154 - 52 )] - 4012
= [ 2006 x 102 ] - 4012
= 204612 - 4012
= 200600
2006 x 102 - 4012
= 204612 - 4012 = 200600
Bài 1 Tính nhanh
a 103^2
b 52×48
c 75^2-50.75+25^2
a) \(103^2=\left(100+3\right)^2=10000+600+9=10609\)
b) \(52\cdot48=\left(50-2\right)\left(50+2\right)=2500-4=2496\)
c) \(75^2-50\cdot75+25^2=\left(75-25\right)^2=50^2=2500\)
\(103^2=\left(100+3\right)^2=100^2+2.3.100+3^2=10000+600+9=10609\)
\(52.48=\left(50+2\right)\left(50-2\right)=50^2-2^2=2500-4=2496\)
\(75^2-50.75+25^2=75^2-25.75-25.75+25^2=75\left(75-25\right)-25\left(75-25\right)=\left(75-25\right)=50^2=2500\)
Tìm x thuộc Z, biết:
a) 1 2 + 1 3 + 1 6 ≤ x ≤ 15 4 + 18 8
b) 11 3 + − 19 6 + − 15 2 ≤ x ≤ 19 12 + − 5 4 + − 10 3
a) 1 ≤ x ≤ 6 ⇒ x ∈ - 3 ; - 2 ; - 1 ; 0 ; 1
b) - 7 ≤ x ≤ - 3 ⇒ x ∈ - 3 ; - 2 ; - 1 ; 0 ; 1
Tìm x, biết:
a) x : 4 5 + 5 2 = − 3 3 4
b) x − 3 4 = 1 5 4
a ) x : 4 5 + 5 2 = − 3 3 4 ⇒ x : 4 5 = − 3 3 4 − 5 2 ⇒ x : 4 5 = − 11 2 ⇒ x = − 22 5 .
b ) x − 3 4 = 1 5 4 ⇒ x − 3 4 = 9 4 ⇒ x = 3 x − 3 4 = − 9 4 ⇒ x = − 3 2 .
so sánh 43/52 và 60/120 , So sánh 17/ 68 và 35 / 103 , So sánh 2018 x 2019-1/2018x2019 va 2019x2020-1/2019x2020
a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)