a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)

a: ĐKXĐ: x<>1; x<>-1

b: \(A=\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-4}{x-1}\)

c: Để A là số nguyên thì x-1-3 chia hết cho x-1

=>\(x-1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;0;4;-2\right\}\)

a: Khi x=5 thì A=5/(5+3)=5/8

b: \(C=A+B=\dfrac{x}{x+3}+\dfrac{2}{x-3}+\dfrac{3-5x}{x^2-9}\)

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

c: Để C nguyên thì x+3-6 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;-9\right\}\)

1) A = \(\dfrac{2x-1}{x+3}\) = \(\dfrac{3}{2}\) (=) (2x-1).2 = 3.(x+3)

                          (=) 4x-2 =3x+9

                          (=) 4x-3x = 9+2

                         (=) x = 11 (tm)

2) Để \(\dfrac{A}{B}\)< \(^{x^2}\)+5 (=) \(\dfrac{2x-1}{x+3}\): \(\dfrac{2}{x^2-9}\) <  \(x^2\)+5 

                    (=) \(\dfrac{\left(2x-1\right)}{\left(x+3\right)}.\dfrac{\left(x-3\right)\left(x+3\right)}{2}\) < \(x^2\)+5

                    (=) \(\dfrac{\left(2x-1\right).\left(x-3\right)}{2}< x^2+5\)

                    (=) \(\dfrac{2x^2-6x-x+3}{2}\) < \(x^2\) +5

                    (=) \(2x^2\)- 7x + 3 < \(2x^2\)+ 10

                    (=)  (\(2x^2\)-\(2x^2\)) - 7x < -3 +10

                    (=) -7x < 7 

                    (=) x > -1

A(x)=(1-x^n)(1+x^n)/(1-x)(1+x)

B(x)=1-x^n/1-x

A(x) chia hết cho B(x) khi 1-x^n chia hết cho 1+x

x^n+1/x+1=A(x)+(1+(-1)^n)/(x+1)

=>1-x^n chia hết cho 1+x khi và chỉ khi n=2k+1