\(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\) (1)

a) ĐKXĐ: \(x\ne\pm3\)

b) \(\left(1\right)=\left[\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{4}{5x+15}\)

\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5x+15}\)

\(=\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)

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

\(=\dfrac{4}{5\left(x-3\right)}\)

c) Thay \(x=19\) vào \(A=\dfrac{4}{5\left(x-3\right)}\) ta có:

\(A=\dfrac{4}{5.\left(19-3\right)}=\dfrac{4}{80}=\dfrac{1}{20}\)

Vậy \(x=19\) thì \(A=\dfrac{1}{20}\)

a) ĐK: \(x\)≠\(+-3\)

b) \(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\)

\(=\dfrac{2x^2+3\left(x+3\right)-x\left(x-3\right)}{x^2-9}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x+3\right)\left(x-3\right)}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4\left(x+3\right)^2}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4}{5\left(x-3\right)}=\dfrac{4}{5x-15}\)

c) Tại x=19

⇒ \(A=\dfrac{4}{5.19-15}=\dfrac{4}{80}=\dfrac{1}{20}\)

Vậy ...

=x2+6x+9(x−3)(x+3).45(x+3)=x2+6x+9(x−3)(x+3).45(x+3)

=45(x−3)=45(x−3)

c) Thay  vào A=45.(19−3)=480=120A=45.(19−3)=480=120

Vậy  thì 

a: ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

b: \(P=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)​

a) \(\frac{x+5}{x^2+5x-6}\)

ĐKXĐ: \(x^2+5x-6=x^2+6x-x-6\)

                                       \(x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)

            \(\Rightarrow x\ne1\) và \(x\ne-6\)                                                            

b) \(\frac{1-x}{3x^2-7x-6}\)

ĐKXĐ: \(3x^2-7x-6=3x^2-9x+2x-6=\left(x-3\right)\left(3x+2\right)\)

\(\Rightarrow x\ne3\) và \(x\ne\frac{-2}{3}\)

Điều kiện để phân thức xác định là x - 5 ≠ 0 ⇔ x ≠ 5

ĐKXĐ:
\(x-2>0\Rightarrow x>2\)

ĐKXĐ: x>2

a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5

b: \(M=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}\)

\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)

a) Điều kiện xác định của phân thức A là x#+-5
\(A=\frac{2\left(x+15\right)}{x^2-25}-\frac{x+3}{x+5}+\frac{x}{x-5} \)
\(A=\frac{2\left(x+15\right)}{\left(x+5\right)\left(x-5\right)}-\frac{x+3}{x+5}+\frac{x}{x-5}\)
\(A=\frac{2\left(x+15\right)}{\left(x+5\right)\left(x-5\right)}-\frac{\left(x+3\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{x\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{2x+30-\left(x^2-5x+3x-15\right)+x^2+5x}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{2x+30-x^2+5x+3x-15+x^2+5x}{\left(x+5\right)\left(x-5\right)}=\frac{15x+15}{\left(x+5\right)\left(x-5\right)}=\frac{15\left(x+1\right)}{\left(x+5\right)\left(x-5\right)}\)

tick đúng nha, ý b tí mình giải nhé