Question

a+√2=√13

二次根式的运算
已知a=(2√2+√5)/(√5-√2)，求代数式a^5-7a^4+6a^3-7a^2+11a+13的值．

Answer 1

a=(2√2+√5)/(√5-√2)
=(2√2+√5)(√5+√2)/(√5-√2)(√5+√2)
=(2√10+4+5+√10)/(5-2)
=(3√10+9)/3
=√10+3
a-3=√10
a^2-6a+9=10
a^2-6a=1
a^5-7a^4+6a^3-7a^2+11a+13
=a^5-6a^4-a^4+6a^3-7a^2+11a+13
=a^3(a^2-6a)-a^2(a^2-6a)-7a^2+11a+13
=a^3*1-a^2*1-7a^2+11a+13
=a^3-8a^2+11a+13
=a^3-6a^2-2a^2+12a-a+13
=a(a^2-6a)-2(a^2-6a)-a+13
=a*1-2*1-a+13
=11