Bài 4:
a: \(A=\dfrac{x+3-x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{\left(x-3\right)^2}{6}=x-3\)
b: Thay x=2003 vào A, ta được:
A=2003-3=2000
b3:
a, 3x(x-1)+x-1=0
(x-1)(3x+1)=0
=>x=1 hoặc x=\(-\dfrac{1}{3}\)
b,
\(4x^2+y^2-20x-2y+26=0\\ 4x^2+y^2-20x-2y+25+1=0\\ \left(4x^2-20x+25\right)+\left(y^2-2y+1\right)=0\\ \left(2x-5\right)^2+\left(y-1\right)^2=0\)
\(\left\{{}\begin{matrix}2x-5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=1\end{matrix}\right.\)
b4:
a,
\(A=\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right):\dfrac{6}{x^2-6x+9}\\\left(\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6}{x^2-6x+9}\)
\(=\dfrac{x+3-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{6}{x^2-6x+9}\\ =\dfrac{6}{\left(x-3\right)\left(x+3\right)}:\dfrac{6}{\left(x-3\right)\left(x-3\right)}=\dfrac{6\left(x-3\right)\left(x-3\right)}{6\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x-3}{x+3}\)
b, thay x=2003 vào bt A ta được :
\(A=\dfrac{x-3}{x+3}=\dfrac{2003-3}{2003+3}=\dfrac{2000}{2006}=\dfrac{1000}{1003}\)
