407 - [ (190 - 170 ): 5 + 9] : 2

407 - [ 20 : 5 + 9]:2

407 - [ 4 + 9]:2

= 407 - 6,5

= 400,5

407 - [( 190 - 170 ) : 4 + 9 ] : 2

= 407 - [ 20 : 4 + 9 ] : 2

= 407 - [ 5 + 9 ] : 2

= 407 - 14 : 2

= 407 - 7

= 400

B)TK:

+)(23.36-17.36):36

=[36.(23-17)]:36

=36.6:36

=(36:36).6
=1.6
=6

b: =23-17=6

a: \(=32+140:70-49=32+2-49=34-49=-15\)

b: \(=408-\left[20:4+9\right]:2=408-7=401\)

c: \(=-564-724+564+224=-500\)

36/20+15/20= 51/20

7/9-6/9= 1/9

\(\dfrac{9}{5}+\dfrac{3}{4}=\dfrac{36}{20}+\dfrac{15}{20}=\dfrac{36+15}{20}=\dfrac{51}{20}\)

\(\dfrac{7}{9}-\dfrac{2}{3}=\dfrac{7}{9}-\dfrac{6}{9}=\dfrac{1}{9}\)

\(a,=\dfrac{x+3+4x-3}{xy}=\dfrac{5x}{xy}=\dfrac{5}{y}\\ b,=\dfrac{x-2+4x-3}{x-1}=\dfrac{5\left(x-1\right)}{x-1}=5\\ c,=\dfrac{x^2+4-6x+5}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\\ d,=\dfrac{4-x^2+2x^2-2x+5-4x}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)

a) \(\dfrac{5xy^2-x^2y+4xy^2+5x^2y}{3xy}=\dfrac{9xy^2+4x^2y}{3xy}=\dfrac{xy\left(9y+4x\right)}{3xy}=\dfrac{9y+4x}{3}\)

b) \(\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=\dfrac{3\left(x+2\right)}{x+2}=3\)

\(\dfrac{5xy^2-x^2y}{3xy}+\dfrac{4xy^2+5x^2y}{3xy}=\dfrac{5xy^2-x^2y+4xy^2+5x^2y}{3xy}=\dfrac{9xy^2+4x^2y}{3xy}=\dfrac{xy\left(9y+4x\right)}{3xy}=\dfrac{9y+4x}{3}\)

\(\dfrac{x+3}{x+2}+\dfrac{x-1}{x+2}+\dfrac{x+4}{x+2}=\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=\dfrac{3\left(x+2\right)}{x+2}=3\)

kết quả thôi nha không nó dài lắm:
e) =\(\dfrac{3}{2x}\)
f) =\(\dfrac{2x}{\left(x+2\right)\left(x-2\right)}\)
g) =\(\dfrac{2}{x+1}\)
h) =\(\dfrac{4x+5}{x+2}\)