1.若x+y=3,xy=1,求(x-y)²,x²+y²的值.

8个回答

  • 1 (x-y)²=(x+y)^2-4xy=3^2-4*1=5,

    x²+y²=(x+y)^2-2xy=3^2-2*1=7

    2 a²b+ab²=(a+b)ab=-30

    又a+b=3,所以ab=-10

    a²+b²=(a+b)^2-2ab=3^2-2*(-10)=29

    3 x²+y²-2x-6y+10=0

    x²-2x+1+y²-6y+9=0

    (x-1)^2+(y-3)^2=0

    所以x-1=y-3=0即x=1,y=3

    4 (3+1)(3²+1)(3^4+1)(3^8+1)+1/2

    =1/2(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)+1/2

    =1/2(3^2-1)(3^2+1)(3^4+1)(3^8+1)+1/2

    =1/2(3^4-1)(3^4+1)(3^8+1)+1/2

    =1/2(3^8-1)(3^8+1)+1/2

    =1/2(3^16-1)+1/2

    =1/2*3^16

    5(1-1/2^2)=(1-1/2)(1+1/2)=(1/2)*(3/2)

    (1-1/3^2)=(1-1/3)(1+1/3)=(2/3)*(4/3)

    .

    (1-1/10^2)=(1-1/10)(1+1/10)=(9/10)*(11/10)

    所以,(1-1/2²)(1-1/3²)(1-1/4²).(1-1/9²)(1-1/10²)=1/2 * 11/10=11/20

    6 如果x²-ax+1是一个整式的平方,则常数a的值为2;如果x²-x+a是一个整式的平方,则常数a的值为1/4