Question

加油~~

想知道兩題的算法

3-2.設x+y=3,xy=1,a+b=4,ab=2,則:
(1)(ax+by)2+(bx-ay)²的值為【
(2)(ax+by)(bx+ay)的值為【 22 】

Answer 1

✨ 最佳解答 ✨

qn

(1)
(ax+by)² + (bx-ay)²
= (a²x² + b²y² + 2abxy) + (b²x² + a²y² - 2abxy)
= a²x² + b²y² + b²x² + a²y²
= (a²+b²)x² + (a²+b²)y²
= (a²+b²)(x²+y²)

a²+b² = (a²+b²+2ab) - 2ab
= (a+b)² - 2·ab
= 4² - 2 · 2
= 12
x²+y² = (x²+y²+2xy) - 2xy
= (x+y)² - 2·xy
= 3² - 2 · 1
= 7

所以 (ax+by)²+(bx-ay)² = (a²+b²)(x²+y²) = 12·7 = 84

(2)
(ax+by)(bx+ay)
= abx² + a²xy + b²xy + aby²
= xy(a²+b²) + ab(x²+y²)
= 1 · 12 + 2 · 7
= 12 + 14
= 26