Question

已知a=1/1-√2,b=1/1＋√2，求代数式a3b＋ab3的值。

求√b2-2b＋1／b2-b的值

Answer 1

a=1/(1-√2)
=(1+√2)/[(1-√2)(1+√2)]
=-(1+√2)

b=1/(1+√2)
=(√2-1)/[(√2-1)(1+√2)]
=√2-1

b-1=√2-1-1=√2-2<0

a^3b+ab^3
=ab(a^2+b^2)
=ab(a^2+b^2+2ab-2ab)
=ab[(a+b)^2-2ab]
=-(1+√2)(√2-1){[-(1+√2)+(√2-1)]^2-2*[-(1+√2)(√2-1)]}
=-{[-1-√2+√2-1]^2-2*[-1]}
=-{[-2]^2+2}
=-{4+2}
=-6

√(b^2-2b+1)／(b^2-b)
=√(b-1)^2／(b^2-b)
=(1-b)/[b(b-1)]
=-(b-1)/[b(b-1)]
=-1/b
=-1/[1/(1+√2)]
=-(1+√2)
=-1-√2

Answer 2

a=1/1-√2=-1-√2 , b=1/1＋√2=√2-1
∴a+b=-2 ab=-1
a3b＋ab3=ab(a²+b²)=ab[(a+b)²-2ab]=-6
√(b2-2b＋1)／(b2-b)=(1-b)/[b(b-1)]=-1/b=-1-√2

Answer 3

ab=-1,a+b=-2
a3b＋ab3=ab(a2+b2)
=ab[(a+b)2-2ab]
=-1[4-2(-1)]
=-6
√b2-2b＋1／b2-b=1-b/b(b-1)=-1/b=-1-√2