\(a,đk\left(B\right):x\ne\pm3\\ B=\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\\ =\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{x^2-9}\\ =\dfrac{3x+9+6x+x^2-3x}{x^2-9}\\ =\dfrac{x^2+6x+9}{x^2-9}\\ =\dfrac{\left(x+3\right)^2}{x^2-9}\\ =\dfrac{x+3}{x-3}\)

\(b,P=A.B\\ =\dfrac{x+1}{x+3}\times\dfrac{x+3}{x-3}\\ =\dfrac{x+1}{x-3}\)

\(c,\) Để P nguyên 

\(\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}\)

=> \(x-3\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{-1;1;2;-2;4;-4\right\}\)

\(=>x=\left\{2;4;5;1;7;-1\right\}\)

Ta có

B   =   a ( a   –   b ) 3   +   2 b ( b   –   a ) 3     =   a ( a   –   b ) 3   –   2 b ( a   –   b ) 3   =   ( a   –   2 b ) ( a   –   b ) 3

Mà a – 2b = 0 nên B   =   0 . ( a   –   b ) 3   =   0

Vậy B = 0

Đáp án cần chọn là: A

Lời giải:

$a-b=3\Rightarrow b=a-3$. Khi đó:

$A=\frac{a-8}{a-3-5}-\frac{4a-(a-3)}{3a+3}=\frac{a-8}{a-8}-\frac{3a+3}{3a+3}=1-1=0$

Theo đề bài : \(a-b=3\Rightarrow a=b+3\).

Thay \(a=b+3\) vào \(A\) ta được : 

\(A=\dfrac{a-8}{b-5}-\dfrac{4a-b}{3a+3}\)

\(=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)

\(=\dfrac{b-5}{b-5}-\dfrac{4b+12-b}{3b+9+3}\)

\(=1-\dfrac{3b+12}{3b+12}=1-1=0\)

Vậy : Với \(a-b=3\) thì \(A=0.\)

\(a-b=3\\ \Rightarrow a=3+b\)

Thay \(a=3+b\) vào \(A\)

\(A=\dfrac{b+3-8}{b-5}-\dfrac{4.\left(b+3\right)-b}{3.\left(b+3\right)+3}\\ =\dfrac{b-5}{b-5}-\dfrac{4b+12-b}{3b+9+3}\\ =\dfrac{b-5}{b-5}-\dfrac{3b+12}{3b+12}\\ =1-1=0\)

Vậy \(A=0\)

B = 2(a³ + b³) - 6(a² + b²) + 3(a² + b²) + (a - b)² + 10ab

= 2(a³ + b³) - 3(a² + b²) + a² + b² - 2ab + 10ab

= 2(a³ + b³) - 2(a² + b²) + 8ab

= 2(a + b)(a² - ab + b²) - 2(a² + b²) + 8ab 

= 4(a² - ab + b²) - 2(a² + b²) + 8ab (Vì a + b = 4)

= 4(a² + b²) - 4ab - 2(a² + b²) + 8ab 

= 2(a² + b²) + 4ab 

= 2(a² + 2ab + b²) 

= 2(a + b)² = 2.4² = 32

Vậy B = 32

vầy b = 32 nhé