a) ĐKXĐ :  \(x\ne-3;x\ne2\)

b) \(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)

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

Để \(A\inℤ\Rightarrow x-3⋮x-2\)

=> \(x-2-1⋮x-2\)

Vì \(x-2⋮x-2\)

=> \(1⋮x-2\)

=> \(x-2\inƯ\left(1\right)\)

=> \(x-2\in\left\{1;-1\right\}\)

=> \(x\in\left\{3;1\right\}\)

Vậy \(x\in\left\{3;1\right\}\)là giá trị cần tìm

a + b , ĐK \(x\ne2;-3\)

\(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5}{\left(x+3\right)\left(x-2\right)}\)

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

Bài làm 

a, \(2x^2\left(3x-5\right)=6x^3-10x^2\)

b, \(\left(12x^3y^2-18x^2y\right):2xy=\left(12x^3y^2:2xy\right)-\left(18x^2y:2xy\right)=6x^2y-9x\)

Bài làm 

\(\left(x^2-2\right)\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

=x²-x³-2+2x+x³+3³

⁼x²-x³-2+2x+x³+27

=x²+25+2x

3( 3x - 5 ) = 9x² - 25

⇔ 3( 3x - 5 ) = ( 3x - 5 )( 3x + 5 )

⇔ 3( 3x - 5 ) - ( 3x - 5 )( 3x + 5 ) = 0

⇔ ( 3x - 5 )[ 3 - ( 3x + 5 ) ] = 0

⇔ ( 3x - 5 )( 3 - 3x - 5 ) = 0

⇔ ( 3x - 5 )( -3x - 2 ) = 0

⇔ 3x - 5 = 0 hoặc -3x - 2 = 0

⇔ x = 5/3 hoặc x = -2/3

3(3x - 5) = 9x² - 25

=> 3(3x - 5) = (3x - 5)(3x + 5)

=> (3x - 5)(3x + 5) - 3(3x - 5) = 0

=> (3x - 5)(3x + 5 - 3) = 0

=> (3x -5)(3x + 2) = 0

=> \(\orbr{\begin{cases}3x-5=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{3};-\frac{2}{3}\right\}\)là giá trị cần tìm

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

b, \(x^3+2-2x^2-x=x\left(x^2-1\right)+2\left(1-x^2\right)\)

\(=x\left(x^2-1\right)-2\left(x^2-1\right)=\left(x-2\right)\left(x^2-1\right)=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)