已知x+y=3,xy=-5.求下列代数式的值 (1)x^2+y^2 (2)x^4+y^4

问题描述:

已知x+y=3,xy=-5.求下列代数式的值 (1)x^2+y^2 (2)x^4+y^4

x^2+y^2=19
x^4+y^4=306.594

X^2+Y^2=(X+Y)^2-2XY=9+10=19
X4+Y4=(X^2+Y^2)^2-2X^2Y^2=(x2+y2)^2-2(xy)^2=311

1)x²+y²=(x+y)²-2xy=9+10=19;
2)原式=(x²+y²)²-2(xy)²=311;

已知x+y=3,xy=-5. 求(1)x^2+y^2 (2)x^4+y^4
(1)x^2+y^2 = (x+y)^2 - 2xy = 9 + 10 = 19
(2)x^4+y^4=(x^2+y^2)^2 - 2x^2y^2 = 19^2 - 2*5^2= 361-50 = 311

(1)x^2+y^2
=(x+y)²-2xy
=9+10
=19
(2)x^4+y^4
=(x²+y²)²-2(xy)²
=361-50
=311

1)x^2+y^2=2(x+y)
因为x+y=3,所以上式=2×3
2)x^4+y^4
与上题解法一样,都是要提取公因式,然后再计算