c: \(=0.5\cdot10-9=5-9=-4\)

c: 

a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)

\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)

\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)

a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)

b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)

d) \(\left(\dfrac{3}{4}+0,5+25\%\right).2\dfrac{2}{3}=\left(\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{4}\right).\dfrac{8}{3}=\left(\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}\right).\dfrac{8}{3}=\dfrac{3}{2}.\dfrac{8}{3}=4\)

a: =3990-(463*72-39^2):15

=3990-2121=1869

b: =1995-963-172=860

c: =33/11-7/11=26/11

d: =3/2+5=13/2

`@` `\text {Answer}`

`\downarrow`

`a,`

`3990 - (463 \times 72 - 39 \times 39) \div 15`

`= 3990 - (33336 - 1521) \div 15`

`= 3990 - 31815 \div 15`

`= 3990 - 2121`

`= 1869`

`b,`

`1995 - 321 \times 3 - 6020 \div 35`

`= 1995 - 963 - 172`

`= 1032 - 172`

`= 860`

`c,`

`3 - 7/11`

`= 33/11 - 7/11`

`= 26/11`

`d,`

`1/4 \div 1/6 + 5`

`= 1/4 \times 6 + 5`

`= 3/2 + 5`

`= 13/2`

Bài 2: 

a: A=(12,87+14,13)+(-14,7-37,3)

=27-52=-25

b: B=-1/3+2/5-2/3-3/5+1/5

=-1

a) 1-3+5-7+9-12+15-18

= (1+ 9) - (3+7) + (5+15) - (12+18) 

= 10 - 10 + 20 - 30 = -10.

b) (-2)+5-7+9-11+13-15+19-21

= 5 - 11 - 2 -2 = 5 - (11+2+2)

= 5 - 15 = -10 

Bài 1: 

a) Ta có: 1-3+5-7+9-12+15-18

=(1+9)-(3+7)+(5+15)-(12+18)

=10-10+20-30

=-10

b) Ta có: \(\left(-2\right)+5-7+9-11+13-15+19-21\)

\(=3+2+2+4-21\)

\(=5+6-21\)

=11-21=-10

Bài 5) Tính giá trị của biểu thức

a) A=1.2+2.3+3.4+...+9.10

b) B=3.4+4.5+5.6+...+198.199+199.200

c) C=1.2.3+2.3.4+3.4.5+...+8.9.10

d) D=31.32.33+32.33.34+...+58.59.60

e) E=1.3+3.5+5.7+...+95.97+97.99

f) F=51.53+53.55+...+153.155+155.157

g) G=1.3.5+3.5.7+...+15.17.19+17.19.21

h) H=2.4+4.6+6.8+...+96.98+98.100

\(a,=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\\ b,=8x^3+27-8x^3+72x=72x+27\)

a) \(=4x^2-4x+1-4\left(x^2-1\right)-\left(x^2-2x+3x-6\right)=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\)

b) \(=8x^3+27-8x\left(x^2-9\right)=8x^3+27-8x^3+72x=72x+27\)

a) (3a + 1)³ = (3a)³ + 3.(3a)².1 + 3.3a.1² + 1³

                   = 27a³ + 27a² + 9a + 1

b) (4 - 2b)³ = 4³ - 3.4².2b + 3.4.(2b)² - (2b)³

                   = 64 - 96b + 48b² - 8b³

a: =1/6+14/6-3/6=12/6=2

b: =-13/8+5/4:(-5/4)

=-13/8-1=-21/8

c: =-3/8(2/5+14/5)

=-3/8*16/5

=-6/5

d: =5/34(1/4+11/9-2/9+29/4)

=5/34*(15/2+1)

=5/34*17/2
=5/4

1:

a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)

b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)

=>2x-1>0

=>x>1/2

2:

a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)

\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)

\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)

\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)

\(=45\sqrt{2}-19\sqrt{5}\)

b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)

\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)

\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)