\(a,=\dfrac{2\left(2x^2+1\right).\left(3x+2\right).2\left(2-x\right)}{\left(x-2\right)\left(x-4\right)\left(2x^2+1\right)}=\dfrac{-4.\left(3x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-4\left(3x+2\right)}{x-4}\\ b,=\dfrac{\left(x+3\right).\left(x+2\right)}{x.\left(x+3\right)^2}\times\dfrac{x\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+3\right)\left(x+2\right)x\left(x+3\right)}{x.\left(x+3\right)^2.\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

a)

\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);                                                   

b)

\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);

c)

\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);                                                               

d)

\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

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

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

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

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

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)

\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)

b: \(=\dfrac{7x-42-x^2+36}{x\left(x-6\right)}=\dfrac{-x^2+7x-6}{x\left(x-6\right)}=\dfrac{-x+1}{x}\)

\(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(x+3\right)-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-3x-9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x}\)

b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)

=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)

=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)

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

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

=\(\dfrac{x+3}{x}\)

#Fiona

c) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)

=\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{3^2-x^2}\) + \(\dfrac{x}{x+3}\)

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

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

=\(\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{9+6x+x^2}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{3^2+2.3x+x^2}{\left(x-3\right)\left(x+3\right)}\)

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

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

=\(\dfrac{x-3}{x+3}\)

#Fiona 

Tick đúng giúp mình nhaa<3

d)\(\dfrac{5x+10}{4x-8}\).\(\dfrac{x-2}{x+2}\)

=\(\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\) . \(\dfrac{x-2}{x+2}\)

=\(\dfrac{5\left(x+2\right).\left(x-2\right)\text{​​}\text{​​}}{4\left(x-2\right).\left(x+2\right)}\)

=\(\dfrac{5}{4}\)

#Fiona

Tick đúng giúp mikk nhaa

\(=\dfrac{x^2+2x+1+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-x-9}{\left(x-2\right)\left(x+2\right)}\)

= 1/48

a) 1/8 x 1/6 = 1/48

1/48

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

\(=-\dfrac{x}{x^2+x+1}\)

`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?

`b,`

\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)

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