Question

A^2b^2+a^2+b^2+1=4ab,求a,b的值 9(x-2)^2-(x-1)^2,求x的值

Answer 1

a^2b^2+a^2+b^2+1=4ab

a^2b^2+a^2+b^2+1-4ab=0

a^2b^2-2ab+1+a^2-2ab+b^2=0

(ab-1)^2+(a-b)^2=0

(ab-1)^2=0,(a-b)^2=0

a=b

a^2=1

a=±1

所以a=b=±1

9(x-2)^2-(x-1)^2=0

[3(x-2)-(x-1)][3(x-2)+(x-1)]=0

(3x-6-x+1)(3x-6+x-1)=0

(2x-5)(4x-7)=0

x=5/2或x=7/4