\(a,27.5^2-25.127=27.25-25.127=25\left(27-127\right)=25.\left(-100\right)=-2500\\ b,\dfrac{-5}{12}+\dfrac{3}{4}+\dfrac{1}{-3}=\dfrac{-5}{12}+\dfrac{9}{12}-\dfrac{4}{12}=\dfrac{0}{12}=0\)

a = -2500

b = 0

\(a,=\dfrac{21}{48}+\dfrac{10}{48}=\dfrac{31}{48}\\ b,=\dfrac{44}{60}-\dfrac{25}{60}=\dfrac{19}{60}\)

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

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: =-12/20+5/20-6/20

=-13/20

b: =10/50-45/50-14/50

=-49/50

c: =31/23-7/32-8/23=1-7/32=25/32

`a)-3/5 + 1/4 + -3/10= (-12)/20 + 5/20 + (-6)/20 = -13/20` 

`b)1/5 + -9/10 + (-7/25)= 10/50 + (-45)/50 + (-14)/50 = (-49)/50`

`c)31/23 - ( 7/32 + 8/23 )=  31/23 - 417/736 =992/736 - 417/736  =  575/736=25/32  `

a)

\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-12\times\dfrac{144}{1}\)

\(=-1728\)

b)

\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)

\(=3-17\)

\(=-14\)

a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)

   

a, \(A=2\sqrt{3}-\sqrt{12}-\sqrt{9}\)

\(=2\sqrt{3}-2\sqrt{3}-3=-3\)

b, \(B=\sqrt{3}\left(\sqrt{12}+\sqrt{27}\right)\)

\(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}\right)\)

\(=\sqrt{3}.5\sqrt{3}=5.3=15\)

a. 7/9 - 16/9 = -9/9 = -1

b. 2/-15 + 7/10 = 17/30

c. (4 2/3 - 4 3/4) : -5/12 - 4/5 

= (14/3 - 19/4) : (-5/12) - 4/5

= -1/12 : (-5/12) - 4/5

= 1/5 - 4/5

= -3/5

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

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

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

Bài 2

A =  (x + 2)^ 2 - x - 3 x (x + 1) = x² + 4x + 4 - x² + 2x + 3 = 6x + 7

B = x^3 - 2x² + 5x - 10 = x² x (x - 2) + 5 x (x - 2) = (x - 2) x (x² + 5)

Vậy x^3 - 2x² + 5x - 10 : (x - 2) = x² + 5