9-12-1+9=
Tập hợp D là?
A. D = {8; 9; 10; 12}
B. D = {1; 9; 10}
C. D = {9; 10; 12}
D. D = {1; 9; 10; 12}
Tập hợp D là?
A. D = {8; 9; 10; 12}
B. D = {1; 9; 10}
C. D = {9; 10; 12}
D. D = {1; 9; 10; 12}
a)2/9-(-4/9)=
b)(-1/24+0,25+1 1/12):(-1/2)2
c)3/13.-5/9+10/13.-5/9+-4/9
Bài 2:
a)-4 1/5:x=5 1/4, b)7/12-5/12.x=1/12
a)2/9-(-4/9)=6/9=2/3
b)(-1/24+0,25+1 1/12):(-1/2)2
=(-1/24+6/24+26/24):(-1/2).2
=31/24.-2.2
=-31/12.2
=-31/6
c)3/13.(-5/9)+10/13.(-5/9)+(-4/9)
=-5/9(3/13+10/13)+(-4/9)
=-5/9.1+(-4/9)
=-1
1.
a)2/9-(-4/9)=6/9
b)(-1/24+0,25+11/12):(-1/2)2
=1,125:-1
=-1,125
c) 3/13.-5/9+10/13.-5/9+-4/9
=-5/9.(3/13+10/13)+4/9
=5/9.1+4/9
=5/9+4/9
=9/9=1
2.
a)-41/5:x=51/4
x=51/41.-41/5
x=-1
vậy x=-1
b)7/12-5/12.x=1/12
5/12.x=1/12+7/12
5/12.x=6/12
x=6/12:5/12
x=6/12.12/5
x=6/5
vậy x=6/5
1- 7/9=.......
7/9+9/12=.......
12:2/3=........
9/12:3=.....
1 - 7/9 = 2/9
7/9 + 9/12 = 55/36
12 : 2/3 = 18
9/12 : 3 = 1/4
\(1-\frac{7}{9}=\frac{9}{9}-\frac{7}{9}=\frac{2}{9}\)
\(\frac{7}{9}+\frac{9}{12}=\frac{28}{36}+\frac{27}{36}=\frac{55}{36}\)
\(12:\frac{2}{3}=12x\frac{3}{2}=\frac{36}{2}=18\)
\(\frac{9}{12}:3=\frac{9}{12}x\frac{1}{3}=\frac{9}{36}=\frac{1}{4}\)
1 - 7/9 = 2/9
7/9 + 9/12 = 55/36
12 : 2/3 = 18
9/12 : 3 = 1/4
1+1+3+9+8+4+6+7+4+5+7+9+9+9+9+9-12
( 1/2 +1 ) .( 1/3 +1 ) .(1/4+1) ........(1/99+1)
9/13 + 9/16 - 9/23
12/13 +12/16 -12/23
tính
Tính: a) 1/8 - 1/2 b) -11/12 - (-2) c) -1/12 - 1/9 d) -5/9 - (-5)/12
a)=\(-\dfrac{3}{8}\)
b)=\(\dfrac{13}{12}\)
c)=\(-\dfrac{7}{36}\)
d)=\(-\dfrac{5}{36}\)
\(\dfrac{1}{8}-\dfrac{1}{2}=\dfrac{1-4}{8}=\dfrac{-3}{8}.\\ \dfrac{-11}{12}-\left(-2\right)=\dfrac{-11}{12}+2=\dfrac{13}{12}.\\ \dfrac{-1}{12}-\dfrac{1}{9}=\dfrac{-7}{36}.\\ \dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{36}.\)
B= 9/10!+ 9/11! + 9/12! + ...+ 9/100! < 1/9!
1/9 - 9/9 x 12/6 + 9/8 - 2/3 =
???????????????????????//
Bài 1 :Cho S =9/10! + 9/11!+ 9/12! +...+ 9/100!. Chứng minh rằng:S<1/9!
12-12+ 11+10-9+8-7+5-4+3-1
3x+ 27=9
2x+12=3(x-7)
2xmũ2 -1 = 49
|-9-x|-5=12