Question

bài 12.cho biểu thức :      \(P=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\).

a,tìm điều kiện xác định của P.

b,rút gọn biểu thức P.

c,tìm giá trị của x để P=4.

Answer 1

\(a,đk:x\ne3;-3\\ b,P=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3\left(x-3\right)+\left(x+3\right)+18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{4}{x-3}\)

\(c,P=4\\ \Rightarrow\dfrac{4}{x-3}=4\\ \Rightarrow4=\left(x-3\right).4\\ \Rightarrow4=4x-12\\ \Rightarrow-4x=-16\\ \Rightarrow x=4\left(thoamanđk\right)\)

 

Answer 2

a. \(x\ne\pm3\)

b. \(=\dfrac{3\left(x-3\right)}{x^2-9}+\dfrac{x+3}{x^2-9}+\dfrac{18}{x^2-9}\)

\(=\dfrac{3x-9+x+3+18}{x^2-9}\)

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

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

c. Với \(P=4\Leftrightarrow\dfrac{4}{x-3}=4\)

\(\Leftrightarrow4x-12=4\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\left(tm\right)\)