Question

Bài : Tính giá trị biểu thức

a, A= \(\dfrac{-5}{13}\) - \(\dfrac{2}{23}\) x a+\(^{b^2}\) tại a= \(\dfrac{-46}{39}\); b = \(\dfrac{-3}{2}\)

b, B= ( \(\dfrac{2}{a}\) + \(\dfrac{b}{-5}\) x \(1\dfrac{3}{7}\) ) : \(\dfrac{3}{-7}\) tại a = \(\dfrac{-2}{3}\) ; b = \(\dfrac{1}{2}\)

Answer 1

a: Thay \(a=-\dfrac{46}{39};b=-\dfrac{3}{2}\) vào A, ta được:

\(A=-\dfrac{5}{13}-\dfrac{2}{23}\cdot\dfrac{-46}{39}+\left(-\dfrac{3}{2}\right)^2\)

\(=\dfrac{-5}{13}+\dfrac{2}{39}\cdot\dfrac{46}{23}+\dfrac{9}{4}\)

\(=-\dfrac{5}{13}+\dfrac{2}{39}\cdot2+\dfrac{9}{4}=\dfrac{-5}{13}+\dfrac{4}{39}+\dfrac{9}{4}\)

\(=\dfrac{-11}{39}+\dfrac{9}{4}=\dfrac{-44+351}{156}=\dfrac{307}{156}\)

b: \(B=\left(\dfrac{2}{a}+\dfrac{b}{-5}\cdot1\dfrac{3}{7}\right):\dfrac{3}{-7}\)

\(=\left(\dfrac{2}{a}-\dfrac{b}{5}\cdot\dfrac{10}{7}\right)\cdot\dfrac{-7}{3}\)

\(=\left(\dfrac{2}{a}-\dfrac{2b}{7}\right)\cdot\dfrac{-7}{3}\)

Thay \(a=-\dfrac{2}{3};b=\dfrac{1}{2}\) vào B, ta được:

\(B=\left(2:\dfrac{-2}{3}-\dfrac{2}{7}\cdot\dfrac{1}{2}\right)\cdot\dfrac{-7}{3}\)

\(=\left(-3-\dfrac{1}{7}\right)\cdot\dfrac{-7}{3}=\dfrac{-22}{7}\cdot\dfrac{-7}{3}=\dfrac{22}{3}\)