F.3 MATHS_1 @ bgdwze3的部落格

標題:

F.3 MATHS

發問:

1) a^2b^2-a^2-b^2+1=?? 2)a+b=10 ,ab=5, a^2+b^2=??

最佳解答:

1) ( ab )^2 - a^2 - b^2 + 1 = [ ( ab )^2 - a^2 ] - ( b^2 -1 ) Grouping = a^2 ( b^2 - 1 ) - ( b^2 -1 ) Taking Out Common Factor = ( a^2 - 1 )( b^2 - 1 ) Taking Out Common Factor again = ( a + 1 )( a - 1 )( b + 1 )( b - 1 ) Factorization using Identity I ====================== 2) From Identity II, we have a^2 + b^2 = a^2 + 2ab + b^2 - 2ab = ( a + b )^2 - 2ab = ( 10 )^2 - 2( 5 ) = 100 - 10 = 90 ==

其他解答:

a^2b^2-a^2-b^2+1 a^2(b^2-1)-(b^2-1) (a^2-1)(b^2-1) (a+1)(a-1)(b-1)(b+1) _________________ a+b=10 ,ab=5, a^2+b^2 (a+b)^2=(10)^2 (a^2+2ab+b^2)=100 a^2+2(5)+b^2=100 a^2+b^2=90|||||1) a^2b^2-a^2-b^2+1 =a^2(b^2-1)-(b^2-1) =(a^2-1)(b^2-1) =(a-1)(a+1)(b-1)(b+1)2)a+b=10 ,ab=5, a^2+b^2 =(a+b)^2-2ab =100-10 =90