Question

(a+b)的平方为8，(a-b)的平方为3，求a的四次方加上b的四次方的值

Answer 1

(a+b)^2=a^2+2ab+b^2=8............①
(a-b)^2=a^2-2ab+b^2=3.............②

①+②，得2a^2+2b^2=11，即a^2+b^2=5.5
①-②，得4ab=5，即ab=2.5
所以a^4+b^4=(a^2+b^2)^2-2a^2b^2=5.5^2-2*2.5^2=17.75

Answer 2

(a+b)²=a²+b²+2ab=8……①
(a-b)²=a²+b²-2ab=3……②
①-②得：4ab=5，即ab=5/4
(a+b)²(a-b)²
=(a²-b²)²
=a^4+b^4-2(ab)²
=a^4+b^4-2*(5/4)²
=a^4+b^4-25/8=24
即a^4+b^4=24+25/8=27又1/8

Answer 3

由题意：（a+b)²=8→a²+b²+2ab=8 ⑴ (a-b)²=3→a²+b²-2ab=3　⑵
因此：⑴+⑵=2a²+2b²=11→a²+b²=5.5 ⑴-⑵=4ab=5→ab=1.25→2a²b²=2×1.25²=3.125
则：所提问题=(a²+b²)²-2a²b²=5.5²-3.125=27.125

Answer 4

(a+b)²=a²+b²+2ab=8……①
(a-b)²=a²+b²-2ab=3……②
①-②得：4ab=5，即ab=5/4
(a+b)²(a-b)²
=(a²-b²)²
=a^4+b^4-2(ab)²
=a^4+b^4-2*(5/4)²
=a^4+b^4-25/8=24
即a^4+b^4=24+25/8=27又1/8

Answer 5

2个等式两边互乘，得到a^4-2a^2b^2+b^4=24 2个等式两边互减，得到4ab=5 ab=5/4 a^2b^2=25/16 a^4+b^4=24+25/8 =217/8