Question

Tính giá trị biểu thức:

1) \(A=3x^2+2x-1\) tại \(\left|x\right|=\dfrac{1}{3}\)

2) \(B=3x^2y+6x^2y^2+3xy^2\) tại \(x=\dfrac{1}{2},y=-\dfrac{1}{3}\)

Answer 1

1: TH1: x=1/3

A=3*1/3^2+2*1/3-1

=3*1/9+2/3-1

=1/3+2/3-1=0

TH2: x=-1/3

A=3*(-1/3)^2+2*-1/3-1

=3*1/9-2/3-1

=1/3-2/3-1=-4/3

2:\(B=3\cdot\left(\dfrac{1}{2}\right)^2\cdot\dfrac{-1}{3}+6\cdot\left(\dfrac{1}{2}\cdot\dfrac{-1}{3}\right)^2+3\cdot\dfrac{1}{2}\cdot\left(-\dfrac{1}{3}\right)^2\)

\(=-\dfrac{1}{4}+6\cdot\dfrac{1}{36}+\dfrac{3}{2}\cdot\dfrac{1}{9}\)

\(=\dfrac{-1}{4}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{-1}{4}+\dfrac{1}{3}=\dfrac{-3+4}{12}=\dfrac{1}{12}\)