=9/2+13/4-11/4+23/10+27/10

=9/2+(13/4-11/4)+(23/10+27/10)

=9/2+1/2+5

=9/2+1/2+10/2

=10

TA có: \(450\%+3\frac14-2,75+2\frac{3}{10}+2,7\)

=4,5+3,25-2,75+2,3+2,7

=4,5+0,5+5

=5+5

=10

1. \(450\%+3=\dfrac{9}{2}+3=\dfrac{15}{2}\)

2. \(\dfrac{1}{4}-2,75+2=0,25-2,75+2=-0,5\)

3. \(\dfrac{3}{10}+2,7=0,3+2,7=3\)

Câu a:

3^2.10^3 - [13^2 - (5^2.4 + 2^2.15)].10^3

= 10^3.(3^2 - 13^2 + 5^2.4 + 2^2.5)

= 10^3.(9 - 169 + 100 + 60)

= 10^3.(-160 + 100 + 60)

= 10^3.(-60 + 60)

= 10^3.0

= 0

Câu b:

600: {450 :[450 - (4.5^3 - 2^3.5^2)]}

= 600 : {450 :[450 - (4.125 - 8.25)]}

= 600 :{450 : [450 - (500 - 200)]}

= 600:{450 :[450 - 300]}

= 600: {450 : 150]

= 600 : 3

= 200

a,  y + y : 0, 5 + y : 0,25 = 1,05 

     y x 1 + y x 2 + y x 4 = 1,05

     y x ( 1 + 2 + 4 )        = 1,05 

     y x         7                 =  1,05

    y                                = 1,05 : 7 

    y                                =    0,15

D

D

a) \(x+\frac{1}{5}-\frac{2}{7}=4\frac{1}{2}\)

=> \(x+\frac{1}{5}-\frac{2}{7}=\frac{9}{2}\)

=> \(x=\frac{9}{2}+\frac{2}{7}-\frac{1}{5}=\frac{321}{70}\)

b) \(1\frac{3}{4}+x-\frac{7}{8}=5\)

=> \(\frac{7}{4}+x-\frac{7}{8}=5\)

=> \(\frac{7}{4}+x=\frac{47}{8}\)

=> \(x=\frac{33}{8}\)

c) \(\frac{1}{2}-\frac{1}{3}\cdot x=\frac{3}{4}\)

=> \(\frac{1}{3}\cdot x=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}\)

=> \(x=\left(-\frac{1}{4}\right):\frac{1}{3}=\left(-\frac{1}{4}\right)\cdot3=-\frac{3}{4}\)