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) \(x+2y+\left(x-y\right)\)

\(=x+2y+x-y\)

\(=2x+y\)

b) \(2x+y-\left(3x-5y\right)\)

\(=2x+y-3x+5y\)

\(=-x+6y\)

c) \(3x^2-4y^2+6xy+7+\left(-x^2+y^2-8xy+9x+1\right)\)

\(=3x^2-4y^2+6xy+7-x^2+y^2-8xy+9x+1\)

\(=2x^2-3y^2-2xy+9x+8\)

d) \(4x^2y-2xy^2+8-\left(3x^2y+9xy^2-12xy+6\right)\)

\(=4x^2y-2xy^2+8-3x^2y-9xy^2+12xy-6\)

\(=x^2y-11xy^2+2+12xy\)

Bài 3:

a) \(4x^2+4x+1=\left(2x+1\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

Phần d bài 3 mk chưa lm dc

Bài 3:

a: \(\frac{x}{x-3}+\frac{9-6x}{x^2-3x}\)

\(=\frac{x}{x-3}+\frac{-6x+9}{x\left(x-3\right)}\)

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

b: \(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}\)

\(=\frac{3\left(2x-1\right)}{x}\cdot\frac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{3\cdot3x}{2x+1}=\frac{9x}{2x+1}\)

Bài 2:

a: \(\frac{x^3-x}{3x+3}\)

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

b: \(\frac{x^2+3xy}{x^2-9y^2}=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\frac{x}{x-3y}\)

Bài 1:

a: \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

\(=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\frac{2}{-\left(x-3\right)}=\frac{-2}{2}=-1\)

b: \(\frac{2x}{x-y}-\frac{2y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)

c: \(\frac{x+15}{x^2-9}+\frac{2}{x+3}\)

\(=\frac{x+15}{\left(x-3\right)\left(x+3\right)}+\frac{2}{x+3}\)

\(=\frac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+15+2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}\)

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

d: \(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{y^2+x^2}{y^2-x^2}\)

\(=\frac{x+y}{2\left(x-y\right)}-\frac{x-y}{2\left(x+y\right)}+\frac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2\left(x^2+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2+2xy-y^2+2x^2+2y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\frac{2x^2+4xy+2y^2}{2\left(x-y\right)\left(x+y\right)}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x+y}{x-y}\)

c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

\(=x^2+x-10\)

a) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)

b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)

c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)

d) \(\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-3x^2y+3xy^2-y^3-x^3+y^3=3xy^2-3x^2y\)

Tk :

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)