\(a,9x^2-6x+1=\left(3x-1\right)^2\)
ĐKXD: \(x\ne\dfrac{1}{3}\)
b,\(\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\) Tại x = -8 ta có\(\dfrac{8}{3.8-1}=\dfrac{8}{23}\)
a/ Tự làm được.
b/ \(C=\dfrac{3\cdot\left(-8\right)^2-\left(-8\right)}{9\cdot\left(-8\right)^2-6\cdot\left(-8\right)+1}\)
\(=\dfrac{3\cdot8^2+8}{9\cdot8^2+48+1}\)
\(=\dfrac{3\cdot64+8}{9\cdot64+48+1}\)
\(=\dfrac{192+8}{576+48+1}\)
\(=\dfrac{200}{625}=\dfrac{8}{25}\)
c) \(\dfrac{3x^2-x}{9x^2-6x+1}\)
\(=\dfrac{x\cdot\left(3x-1\right)}{\left(3x-1\right)^2}\)
\(=\dfrac{x}{3x-1}\)
