Question

一道无理方程的题目,

Answer 1

√(x^2+3x+4)=√(x^2+3x+2)+1

[√(x^2+3x+4)]^2=[√(x^2+3x+2)+1]^2

x^2+3x+4=x^2+3x+2+2√(x^2+3x+2)+1

x^2+3x+4=x^2+3x+3+2√(x^2+3x+2)

1=2√(x^2+3x+2)

1^2=[2√(x^2+3x+2)]^2

1^2=4(x^2+3...