\(=\dfrac{8}{12}+\dfrac{6}{12}-\dfrac{9}{12}=\dfrac{14}{12}-\dfrac{9}{12}=\dfrac{5}{12}\)
\(=\dfrac{8}{12}+\dfrac{6}{12}-\dfrac{9}{12}=\dfrac{14}{12}-\dfrac{9}{12}=\dfrac{5}{12}\)
a,1/3 .(x-2/5)=3/4 b, 7/3:(x-2/3)=4/5 c,1/3.(x-2/5)=4/5 d, 2/3.(x-1/2)-1/4.(x-2/5)=7/3 e,3/7 .(x-2/3)+1/2=5/4.(x-2) f,1/2.(x-3)+1/3.(x-4)+1/4.(x-5)=1/5 g,[2/3.(x-1/2)-4/5]:(x-1/3)=21/5 h, {x-[1/2.(x-3)+11/5]}:(x-1/2)=3/5 i,x.(x-2/5)-(x+2).x+11/4=4/3
bài 3:
a) x - 3/4 = 6 x 3/8 b) 7/8 : x = 3 - 1/2 c) x + 1/2 x 1/3 = 3/4
d) 3/2 x 4/5 - x = 2/3 e) X x 3 1/3 = 3 1/3 : 4 1/4 f) 5 2/3 : x = 3 2/3 - 2 1/2
1. 3/2 x 4/5 - x = 2/3
2.x x3 1/2=3 1/3 : 4 1/4
3. 5 2/3:x = 3 2/3- 2 1/2
4. 3/2+1/2:x=3 1/2
5.3/4 x x - 1/5=1/5:1/8
Chứng minh rằng:
a,A=1/2+1/2^2+1/2^3+.+1/2^2<1
b,B=1/3+1/3^2+1/3^3+...+1/3^n<1/2
c,B=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2015-1/2^2016<1/3
d,D=1/3+2/3^2+3/3^3+4/3^4+...+100/3^100<3/4
Cho dãy : 1/1 ; 2/1 ; 1/2 ; 3/1 ; 2/2 ; 1/3 ; 4/1 ; 3/2 ; 2/3 ; 1/4 ; 5/1 ; 4/2 ; 3/3 ; 2/4 ; 1/5 ;... . Tìm số thứ 2001 của dãy
a, 4×(1/4)^2+25×[(3/4)^3:(5/4)^3] : (3/2)^3.
b,2^3+3×(1/2)^0-1+[(-2)^2:1/2] - 8
c,(-1/3)^2-(-3/5)^0+(1/2)^2 :2
mn oi ai on hok giúp mik vói
giải cthe cho mik nha .
1+1+1+1+1+1+2+2+2+2+2+2+3+3+3+4+4+4+4+4+4+4+4+1234567890 có chia hết cho 3 không?
1) 1/3 . x + ( 2/3 - 4/9 ) = -3/4
2) 2.x/3 - 1/3 = -3/2 : 7/4
3) 3/4 + 1/4 : x = 2/5
1. (1+1/2).(1+1/2^2).(1+1/2^3)....(1+1/2^100) < 3
2. 1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+ 1024/(5^1024+1) <1/4
3. 3/(1!+2!+3!)+4/(2!+3!+4!)+...+100/(98!+99!+100!) <1/2
b) 1 - 1/2 + 2 - 2/3 + 3 - 3/4 +4 - 1/4 - 3 - 1/3 - 2 - 1/2 -1