若a+b=5,ab=6,求a^2+b^2; a^4+b^4的值

来源：百度知道 编辑：UC知道 时间：2024/07/06 15:54:11

a^2+b^2=(a+b)^2-2ab
=5^2-2*6
=13

a^4+b^4
=(a^2+b^2)^2-2a^2b^2
=(a^2+b^2)^2-2(ab)^2
=13^2-2*6^2
=97

a+b=5
(a+b)^2=a^2+2ab+b^2=25
a^2+b^2=25-2ab=25-2*6=13

a^2+b^2=13
(a^2+b^2)^2=a^4+2a^2b^2+b^4=169
a^4+b^4=169-2a^2b^2
=169-2*(ab)^2
=169-2*36
=97

a^2+b^2=(a+b)^2-2ab=13
a^4+b^4=(a^2+b^2)^2-2a^2b^2=97

a^2+b^2=(a+b)^2-2ab
=13
a^4+b^4=(a^2+b^2)^2-2a^2b^2
=13^2-2*6^2
=97

a^2+b^2=(a+b)^2-2ab=13
a^4+b^4=(a^2+b^2)^2-2(ab)^2=13^2-2*6^2=97;

a+b=5
(a+b)^2=a^2+2ab+b^2=25
a^2+b^2=25-2ab=25-2*6=13

a^2+b^2=13
(a^2+b^2)^2=a^4+2a^2b^2+b^4=169
a^4+b^4=169-2a^2b^2
=169-2*(ab)^2
=169-2*36
=97

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