Question

bài 4 tính nhanh

a. 77^2 +23^2 +77.46

b.105^2-5^2

c. A= (x-y)(x^2+xy +y^2) + 2y^3 tại x = 2/3 và y = 1/3

Answer 1

a) \(77^2+23^2+77\cdot46\)

\(=77^2+2\cdot77\cdot23+23^2\)

\(=\left(77+23\right)^2\)

\(=100^2\)

\(=10000\)

b) \(105^2-5^2\)

\(=\left(105-5\right)\left(105+5\right)\)

\(=100\cdot110\)

\(=11000\)

c) \(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)

\(\Rightarrow A=x^3+y^3+2y^3\)

\(\Rightarrow A=x^3+3y^3\)

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

\(A=\left(\dfrac{2}{3}\right)^3+3\cdot\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow A=\dfrac{8}{27}+\dfrac{3}{27}\)

\(\Rightarrow A=\dfrac{11}{27}\)