分式整式数学题

(2a-b-c)^2+(c-b)^2
=[(a-b)+(a-c)]^2+[(a-b)-(a-c)]^2
=(2+1)^2+(2-1)^2
=10

a-b=(1/20x+20 )-(1/20x+19)=1
b-c=(1/20x+19)-(1/20x+21)=-2
c-a=(1/20x+21)-(1/20x+20)=1
a^2+b^2+c^2-ab-bc-ca
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)]/2
=[(a-b)^2+(b-c)^2+(c-a)^2]/2
=[1^2+(-2)^2+1^2]/2
=3

4a表示4个a相加

xy=(√2+1)(√2-1)=(√2)^2-1^2=1
x-y=(√2+1)-(√2-1)=2
x^2-5xy+y^2+6
=(x^2-2xy+y^2)-3xy+6
=(x-y)^2-3xy+6
=2^2-3*1+6
=7

1.a-b=2,a-c=1,则(2a-b-c)的平方+(c-b)的平方是多少？

(2a-b-c)^2+(c-b)^2
=[(a-b)+(a-c)]^2+[(a-b)-(a-c)]^2
=(2+1)^2+(2-1)^2
=10

2.已知a=1/20x+20 b=1/20x+19 c=1/20x+21 那么代数式a的平方+b的平方+c的平方-ab-bc-ac为多少？
a-b=(1/20x+20 )-(1/20x+19)=1
b-c=(1/20x+19)-(1/20x+21)=-2
c-a=(1/20x+21)-(1/20x+20)=1
a^2+b^2+c^2-ab-bc-ca
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+