Question

已知a,b均为实数，且b=√(2a+1)/(4a-3)+√(1+2a)/(3-4a) +1，则a^2+b^2的值是？

已知a,b均为实数，且b=√(2a+1)/(4a-3)+√(1+2a)/(3-4a) +1，则a^2+b^2的值是？

Answer 1

b=√[(2a+1)/(4a-3)]+√[(1+2a)/(3-4a)] +1
b=√[(2a+1)/(4a-3)]+√[-(1+2a)/(4a-3)] +1
√[(2a+1)/(4a-3)]>=0
√[-(1+2a)/(4a-3)]>=0
所以(2a+1)/(4a-3)=0
即2a+1=0
a=-1/2
b=1

a^2+b^2
=(-1/2)^2+1^2
=1/4+1
=5/4

Answer 2

由(2a+1)/(4a-3)≥0，以及(1+2a)/(3-4a)≥0，则必有2a+1=0，a=-0.5
于是b=1
于是a^2+b^2=1+0.25=1.25，就是5/4

Answer 3

观察两个根号里,发现它们相差一个符号,都要大于等于零,那么只能里面为零,即a=-1/2或a=3/4,而b=1,所以a方+b方=5/4或25/16.

Answer 4

b=√[(2a+1)/(4a-3)]+√[(1+2a)/(3-4a)] +1
b=√[(2a+1)/(4a-3)]+√[-(1+2a)/(4a-3)] +1
√[(2a+1)/(4a-3)]>=0
√[-(1+2a)/(4a-3)]>=0
所以(2a+1)/(4a-3)=0
即2a+1=0
a=-1/2
b=1

a^2+b^2
=(-1/2)^2+1^2
=1/4+1
=5/4