rút gọn phân thức x^2-9/x+3
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Câu 4: Không có nghĩa khi x-3=0
=>x=3
Câu 5:
\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)
\(\dfrac{x^2-3x}{x^2-6x+9}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2}=\dfrac{x}{x-3}.\)
ĐKXĐ: \(x\ne3.\)
\(=\dfrac{\left(x-3\right)\cdot x}{\left(x-3\right)^2}=\dfrac{x}{x-3}\)
\(=\dfrac{x^2-6x+9-x^2-9}{x\left(x-3\right)}=\dfrac{-6x}{x\left(x-3\right)}=\dfrac{6}{3-x}\)
\(2,=x^2-3^2=\left(x-3\right)\left(x+3\right)\\ 3,=\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x=4xy\)
Bài làm
\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)
\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)
\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)
\(a,\) Điều kiện xác định: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
\(b,A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)
\(=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4}{x-3}\)
\(c,x=1\Rightarrow A=\dfrac{4}{1-3}=-2\)
\(=\dfrac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3\)