Question

help

Answer 1

a, đk : x khác 3 ; -3 

\(P=\dfrac{2x}{x-3}+\dfrac{x}{x+3}-\dfrac{3x^2+9}{x^2-9}=\dfrac{2x^2+6x+x^2-3x-3x^2-9}{x^2-9}\)

\(=\dfrac{3x-9}{x^2-9}=\dfrac{3}{x+3}\)

b, Ta có : \(P=\dfrac{3}{x+3}=3\Rightarrow3=3x+9\Leftrightarrow3x=-6\Leftrightarrow x=-2\)

c, \(\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

x + 3	1	-1	3	-3
x	-2	-4	0	-6

 

Answer 2

Câu 1:

-ĐKXĐ: 

x-3≠0   ⇔x≠3

x+3≠0  ⇔x≠-3

x²-9≠0 ⇔x≠3 và x≠-3

- Rút gọn P:

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

Answer 3

help câu 2 với

 

Answer 4

Câu 2:

1. x²-x+m                     x-2

    x²-2x                        x+1

_____________

        x+m

        x-2

______________

          m+2

x²-x+m chia hết cho x-2 ⇔m+2=0 ⇔m=-2.