x = -5 , y = 2 , z = -1
a . (3x+2y)(3z)
b . xy/2 + yz/4 + xz/5
xy......yz......xz
---- + ---- + ----
2........4........5
小學數學題
2007-08-07 1:12 am
回答 (4)
2007-08-07 1:16 am
✔ 最佳答案
a. Since x = -5, y = 2, z = -1So, (3x + 2y)(3z)
= [3(-5) + 2(2)][3(-1)]
= (-15 + 4)(-3)
= (-11)(-3)
= 33
b. xy / 2 + yz / 4 + xz / 5
= (-5)(2) / 2 + (2)(-1) / 4 + (-5)(-1) / 5
= -5 - 1/2 + 1
= -9/2
參考: Myself~~~
2007-08-07 2:03 am
a. (3x+2y)(3z)
=3z(3x+2y)
=9xz+6yz
b. xy/2 + yz/4 + xz/5
=10xy/20+5yz/20+4xz/20
=(10xy+5yz+4xz)/20
=3z(3x+2y)
=9xz+6yz
b. xy/2 + yz/4 + xz/5
=10xy/20+5yz/20+4xz/20
=(10xy+5yz+4xz)/20
參考: ME
2007-08-07 1:26 am
a.(3x+2y)(3z)
=[3(-5)+(2)(2)][3(-1)]
=-11(-3)
=33
or
9xz+6yz
=9(-5)(-1)+6(2)(-1)
=45-12
=33
b.xy/2 + yz/4 + xz/5
=(-5)(2)/2+(2)(-1)/4+(-5)(-1)/5
=(-5)+(-0.5)+(1)
=-5-0.5+1
=-4.5
=[3(-5)+(2)(2)][3(-1)]
=-11(-3)
=33
or
9xz+6yz
=9(-5)(-1)+6(2)(-1)
=45-12
=33
b.xy/2 + yz/4 + xz/5
=(-5)(2)/2+(2)(-1)/4+(-5)(-1)/5
=(-5)+(-0.5)+(1)
=-5-0.5+1
=-4.5
參考: own
2007-08-07 1:18 am
a . (3x+2y)(3z)
=[3(-5)+2(2)][3(-1)]
=(-15+4)(-3)
=(-11)(-3)
=33
b. xy/2 + yz/4 + xz/5
=(-5)(2)/2+(2)(-1)/4+(-5)(-1)/5
=-5-1/2+1
=-4.5
=[3(-5)+2(2)][3(-1)]
=(-15+4)(-3)
=(-11)(-3)
=33
b. xy/2 + yz/4 + xz/5
=(-5)(2)/2+(2)(-1)/4+(-5)(-1)/5
=-5-1/2+1
=-4.5
收錄日期: 2021-05-03 07:01:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070806000051KK03558