a, Để A xác định
\(\Leftrightarrow\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
\(b,A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)
\(=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)
\(=\dfrac{3\left(x-3\right)+x+3+18}{x^2-9}\)
\(=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)
c, Thay x = 1 vào biểu thức trên ,có :
\(\dfrac{4}{1-3}=\dfrac{4}{-2}=-2\)
Vậy tại x = 1 giá trị biểu thức \(\dfrac{4}{x-3}\) là - 2
\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)(1 )
a ) ĐKXĐ : \(x\ne\pm3\)
b ) Rút gọn :
(1 ) \(\Rightarrow A=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)
c ) Khi x = 1, thì :
\(A=\dfrac{4}{1-3}=-2\)
a.ĐKXĐ: x+3≠0 =>x≠-3 ; x-3≠0 =>x≠3 ; 9-x2≠0 => x≠\(\pm\) 3
Vậy ĐKXĐ là:x≠\(\pm\) 3
b.\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4}{x-3}\) c.Thay x=1 vào bt vừa rút gọn ta được
A=\(\dfrac{4}{1-3}=\dfrac{4}{-2}=-2\)