Question

Answer 1

a) \(A=\left(2x+y\right)\left(2x-y\right)\)

Thay \(x=-2;y=\dfrac{1}{3}\) vào A, ta được:

\(A=\left[2\cdot\left(-2\right)+\dfrac{1}{3}\right]\left[2\cdot\left(-2\right)-\dfrac{1}{3}\right]\)

\(=\left(-4+\dfrac{1}{3}\right)\left(-4-\dfrac{1}{3}\right)\)

\(=-\dfrac{11}{3}\cdot-\dfrac{13}{3}=\dfrac{143}{9}\)

Vậy \(A=\dfrac{143}{9}\) tại \(x=-2;y=\dfrac{1}{3}\).

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b) Ta có: \(x\left(3x-2\right)-3x^2=\dfrac{3}{4}\)

\(\Rightarrow3x^2-2x-3x^2=\dfrac{3}{4}\)

\(\Rightarrow-2x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}:\left(-2\right)=-\dfrac{3}{8}\)

Vậy \(x=-\dfrac{3}{8}\) là giá trị cần tìm.