Question

1.已知x=√3+1,.求代数式（x+1）^2-4（x+1）+4的值.

Answer 1

1.已知x=√3+1,x-1=√3

（x+1）^2-4（x+1）+4

=(x+1-2)^2

=(x-1)^2

=(√3)^2

=3

2. 2a（a+b）-（a+b）^2

=(a+b)(2a-a-b)

=(a+b)(a-b)

=a^2-b^2

=3-5

=-2

3. 已知：a+b=3,ab=2

(1) a^2b+ab^2=ab(a+b)=2*3=6

(2) a^2+b^2=(a+b)^2-2ab=3^2-2*2=5

(3) 1/a^2+1/b^2=(a^2+b^2)/(ab)^2=5/2^2=5/4

希望能帮到你,祝学习进步O(∩_∩)O