a: Thay x=1/3 vào A, ta được:
\(A=3\cdot\left(\dfrac{1}{3}\right)^2-4\cdot\dfrac{1}{3}+1=3\cdot\dfrac{1}{9}-\dfrac{4}{3}+1\)
\(=\dfrac{1}{3}-\dfrac{4}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}=0\)
Thay x=2/9 vào A, ta được:
\(A=3\cdot\left(\dfrac{2}{9}\right)^2-4\cdot\dfrac{2}{9}+1\)
\(=3\cdot\dfrac{4}{81}-\dfrac{8}{9}+1\)
\(=\dfrac{4}{27}-\dfrac{8}{9}+1=\dfrac{7}{27}\)
b: \(B=9a^2-6a+1=\left(3a\right)^2-2\cdot3a\cdot1+1^2=\left(3a-1\right)^2\)
\(\left|a\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}a=\dfrac{1}{2}\\a=-\dfrac{1}{2}\end{matrix}\right.\)
Thay a=1/2 vào B, ta được:
\(B=\left(3\cdot\dfrac{1}{2}-1\right)^2=\left(\dfrac{3}{2}-1\right)^2=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
Thay a=-1/2 vào B, ta được:
\(B=\left(3\cdot\dfrac{-1}{2}-1\right)^2=\left(-\dfrac{3}{2}-1\right)^2=\left(-\dfrac{5}{2}\right)^2=\dfrac{25}{4}\)
c: |x-1|+|y+2|=0
=>\(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Thay x=1 và y=-2 vào C, ta được:
\(C=\dfrac{3\cdot1^2-2\cdot1\cdot\left(-2\right)+5}{1^2+\left(-2\right)}\)
\(=\dfrac{3+4+5}{-1}=-12\)
