一、因式分解下列各式:
1. (X+Y)(Y+Z)(Z+X)+XYZ=?
2. (a^2-1)(b^2-1)-4ab=?
二、a+b=24，ab=36，則√a/√b+√b/√a=?
(請各題解釋詳細，才有可能採納哦!)
國二數學老師，求救啊!?
2015-12-05 11:03 am
回答 (2)
2015-12-05 11:32 am
✔ 最佳答案
一1.
(X+Y)(Y+Z)(Z+X) + XYZ
= [ XZ + Y(X+Y+Z) ](Z+X) + XYZ
= XZ(Z+X) + Y(X+Y+Z)(Z+X) + XYZ
= XZ(X+Y+Z) + Y(X+Y+Z)(Z+X)
= (X+Y+Z)(XY+YZ+ZX) ..... Ans
2.
(a^2-1)(b^2-1) - 4ab
= a^2b^2 - a^2 - b^2 + 1 - 4ab
= ( a^2b^2 - 2ab + 1 ) - ( a^2 + b^2 + 2ab )
= (ab-1)^2 - (a+b)^2
= (ab+a+b-1)(ab-a-b-1) ..... Ans
二
√a/√b + √b/√a
= (√a*√a + √b*√b ) / ( √a * √b )
= (a+b) / √(ab)
= 24 / √36
= 24 / 6
= 4 ..... Ans
2016-01-04 3:04 pm
一
1.
(X+Y)(Y+Z)(Z+X) + XYZ
= [ XZ + Y(X+Y+Z) ](Z+X) + XYZ
= XZ(Z+X) + Y(X+Y+Z)(Z+X) + XYZ
= XZ(X+Y+Z) + Y(X+Y+Z)(Z+X)
= (X+Y+Z)(XY+YZ+ZX)
2.
(a^2-1)(b^2-1) - 4ab
= a^2b^2 - a^2 - b^2 + 1 - 4ab
= ( a^2b^2 - 2ab + 1 ) - ( a^2 + b^2 + 2ab )
= (ab-1)^2 - (a+b)^2
= (ab+a+b-1)(ab-a-b-1)
二
√a/√b + √b/√a
= (√a*√a + √b*√b ) / ( √a * √b )
= (a+b) / √(ab)
= 24 / √36
= 24 / 6
= 4
1.
(X+Y)(Y+Z)(Z+X) + XYZ
= [ XZ + Y(X+Y+Z) ](Z+X) + XYZ
= XZ(Z+X) + Y(X+Y+Z)(Z+X) + XYZ
= XZ(X+Y+Z) + Y(X+Y+Z)(Z+X)
= (X+Y+Z)(XY+YZ+ZX)
2.
(a^2-1)(b^2-1) - 4ab
= a^2b^2 - a^2 - b^2 + 1 - 4ab
= ( a^2b^2 - 2ab + 1 ) - ( a^2 + b^2 + 2ab )
= (ab-1)^2 - (a+b)^2
= (ab+a+b-1)(ab-a-b-1)
二
√a/√b + √b/√a
= (√a*√a + √b*√b ) / ( √a * √b )
= (a+b) / √(ab)
= 24 / √36
= 24 / 6
= 4
收錄日期: 2021-04-18 14:10:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151205030313AAtXbPy