Answer:

a) \(P=\frac{5x-2}{x^2-4}-\frac{3}{x+2}+\frac{x}{x-2}\)

\(=\frac{5x-2}{\left(x-2\right)\left(x+2\right)}-\frac{3x-6}{\left(x+2\right)\left(x-2\right)}+\frac{x+2x}{\left(x-2\right)\left(x+2\right)}\)

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

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

\(=\frac{x+2}{x-2}\)

b) \(\left|x+3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}x+3=5\\x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\text{(Loại)}\\x=-8\end{cases}}\) 

Với \(x=-8\) thì giá trị của biểu thức P

\(P=\frac{-8+2}{-8-2}=\frac{3}{5}\)

c) \(P=\frac{x+2}{x-2}=\frac{x-2+4}{x-2}=1+\frac{4}{x-2}\)

Mà để P nguyên thì \(\frac{4}{x-2}\) nguyên

\(\Rightarrow\left(x-2\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;6;-2\right\}\) mà đề ra \(x\ne\pm2\)

Vậy \(x\in\left\{3;1;4;0;6\right\}\) thì P nguyên

=(x^3+6x^2+12x+8)+y^3

=(x^3+3x^2+3x2^2+2^3)+y^3

=(x+2)^3+y^3

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

=(x+2+y)(x^2+4x+4-xy-2y+y^2)

=(x+2+y)(x^2+y^2-xy+4x-2y+4)

ko bít đâu nha : lớp 5

x³ - 6x² + 12x - 8

= x³ - 2x² - 4x² + 4x + 8x - 8

= (x³ - 2x²) - (4x² - 8x) + (4x - 8)

= x².(x - 2) + 4x.(x - 2) + 4.(x - 2)

= (x - 2).(x² + 4x + 4)

= (x - 2).(x² + 2x + 2x + 4)

= (x - 2).[x.(x + 2) + 2.(x + 2)]

= (x - 2).(x + 2).(x + 2)

= (x - 2).(x + 2)²

những bạn nhầm đề bài r bạn

=33311