Question

CHo phân thức A = 3/x+3 + 1/x-3 - 18/9-x² ( x khác 3 ; x khác -3 )

a Rút gọn A 
b Tìm  x để A = 4

Answer 1

\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(A=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{4}{x-3}\)

Answer 2

a)

\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(A=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{-\left(9-x^2\right)}\)

\(A=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{x^2-3^2}\)

\(A=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{\left(x+3\right).\left(x-3\right)}\)

\(A=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{4}{x-3}\)

b) Thay \(A=4\) vào phân thức \(A\) , ta có:

\(\frac{4}{x-3}=4\)

\(\Leftrightarrow x-3=\frac{4}{4}\)

\(x-3=1\)

\(x=1+3\)

\(x=4\)

Vậy \(x=4\) khi \(A=4\)