Bài làm:

\(A=\frac{4^2}{15}.\frac{5^2}{24}.\frac{6^2}{35}.....\frac{20^2}{399}=\frac{4^2}{3.5}.\frac{5^2}{4.6}.\frac{6^2}{5.7}.....\frac{20^2}{19.21}=\frac{\left(4.5.6.....20\right)\left(4.5.6.....20\right)}{\left(3.4.5.....19\right)\left(5.6.7.....21\right)}=\frac{20.4}{3.21}=\frac{80}{63}\)

Học tốt!!!!

a. 215 - 5 + 10 = 210 + 10 = 220

b. 13 + (-17) = -4

c. 25 . 5 . 2014 . 4 . 2 = (25 . 4) . (5 . 2) . 2014 = 100 . 10 . 2014 = 1000 . 2014 = 2014000

đ. 24.35 + 76.35 - 500 = (24 + 76) . 35 - 500 = 100 . 35 - 500 = 3500 - 500 = 3000

e. 3 + 7 + 11 + 15 + ... + 399 + 403 = 406 . 101 : 2 = 20503

ta có 

A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{19.21}\)

\(=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{21}\right)\)

=\(\frac{4}{7}\)

a) 12 x 4 : 2 = 48 : 2

                  = 24

b) 35 + 15 : 5 = 35 + 3

                  = 38

a. 2205 - (35 . 7)

= 2205 - 245

= 1960

b. 8700 : 25 . 4

= 348 . 4

= 1392

thanks 

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)

Có vài bước mình làm tắc á nha :>