Question

若a^2+2b^2=5,则代数式(3a^2-2ab+b^2)-(a^2-2ab-3b^2)的值为

要过程

Answer 1

(3a^2-2ab+b^2)-(a^2-2ab-3b^2)
解：原式=3a²-2ab+b²-a²+2ab+3b²
=2a²+4b²
=2﹙a²+2b²﹚
把a^2+2b^2=5代入上式：
原式=2×5=10

Answer 2

(3a^2-2ab+b^2)-(a^2-2ab-3b^2)
=3a²-2ab+b²-a²+2ab+3b²
=2a²+4b²
=2(a²+2b²)
=2×5
=10

Answer 3

解：原式=3a^2-a^2-2ab+2ab+b^2+3b^2
=2a^2+4b^2
=2(a^2+2b^2)
而a^2+2b^2=5，
所以原式=2×5=10

Answer 4

因为a^2+2b^2=5
(3a^2-2ab+b^2)-(a^2-2ab-3b^2)=2a^2+4b^2=2(a^2+2b^2)=10

Answer 5

(3a^2-2ab+b^2)-(a^2-2ab-3b^2)
=3a^2-2ab+b^2-a^2+2ab+3b^2
=2a^2+4b^2
=2(a^2+2b^2)
=10

Answer 6

解：(3a^2-2ab+b^2)-(a^2-2ab-3b^2）=2a^2+4b^2=2×(a^2+2b^2)=2×5=10