A= \(\frac{3\left(x-3\right)+\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{18}{\left(9-x^2\right)}\)
A= \(\frac{3x-9+x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{x^2-9}\)
A=\(\frac{3x+x-9+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)
A=\(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
A=\(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
A=\(\frac{4}{\left(x-3\right)}\)
để A=4
=> \(\frac{4}{x-3}=4\)
<=> x-3=1
<=> x=4
a, Rút gọn :
\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
\(A=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{1\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{18}{\left(x+3\right)\left(x-3\right)}\)
\(A=\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)
\(A=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}\)
\(A=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=\frac{4}{x-3}\)
b, Để A = 4
\(\Leftrightarrow\frac{4}{x-3}=4\)
\(\Leftrightarrow4\left(x-3\right)=4\)
\(\Leftrightarrow4x-12=4\)
\(\Leftrightarrow4x=16\)
\(\Leftrightarrow x=4\)
Vậy để a = 4 thì x = 4
