根号下x^4+x^2y^2化简=(x大于等于0)
网友回答
√(x^4+x^2y^2)
=√[x^2(x^2+y^2)]
=x√(x^2+y^2)
======以下答案可供参考======
供参考答案1:
设x=（tanα）^2+1
则√[x+2√(x-1)]+√[x-2√(x-1)]
=√[（tanα）^2+1+2tanα]+√[(tanα）^2+1+2tanα]
=|tanα+1|+|tanα-1|
又1≤x≤2
则√[x+2√(x-1)]+√[x-2√(x-1)]=2
当2≤x√[x+2√(x-1)]+√[x-2√(x-1)]
=2√(x-1)