Question

因式分解（x²+y²）（2x²+2y²-3）-5 急！！！！！！！

Answer 1

(x²+y²)(2x²+2y²-3)-5=2(x²+y²)²-3(x²+y²)-5

=[2(x²+y²-5]*[2(x²+y²)+1]
=(2x²+2y²-5)(2x²+2y²+1)

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Answer 2

(x" + y") (2x" + 2y" - 3) - 5
= (x" + y") [ 2(x" + y") - 3 ] - 5
= 2(x" + y")" - 3(x" + y") - 5
= 2(x" + y")" - 5(x" + y") + 2(x" + y") - 5
= (x" + y") (2x" + 2y" - 5) + (2x" + 2y" - 5)
= (x" + y" + 1) (2x" + 2y" - 5)

或者看
= (x" + y") (2x" + 2y" + 2 - 5) - 5
= (x" + y") (2x" + 2y" - 5) + 2(x" + y") - 5
= (x" + y") (2x" + 2y" - 5) + 2x" + 2y" - 5
= (x" + y" + 1) (2x" + 2y" - 5)

Answer 3

(x²+y²)(2x²+2y²-3)-5
=2(x²+y²)²-3(x²+y²)-5
=[2(x²+y²)-5]*[2(x²+y²)+1]
=(2x²+2y²-5)(2x²+2y²+1)