Question

初中两道数学题，要快

第一道（A的4次方—2A的平方B的2次方+B的4次方）除以（A+B）的平方除以（A—B）
第二道 已知A，Y满足A的2次方+Y的2次方+4分之5=2A+Y，求A+Y分之AY的值。

Answer 1

(A^4-2A^2B^2+B^4))÷(A+B)^2÷(A-B)
=(A^2-B^2)^2÷(A+B)^2÷(A-B)
=[(A^2-B^2)]^2÷(A+B)^2÷(A-B)
=[(A-B)(A+B)]^2÷(A+B)^2÷(A-B)
=(A-B)^2(A+B)^2÷(A+B)^2÷(A-B)
=(A-B)^2÷(A-B)
=A-B

A^2+Y^2+5/4=2A+Y
A^2-2A+1+Y^2-Y+1/4=0
(A-1)^2+(Y-1/2)^2=0
(A-1)^2=0,(Y-1/2)^2=0
A=1,Y=1/2
AY/(A+Y)
=(1*1/2)/(1+1/2)
=(1/2)/(3/2)
=1/3