一條中二數學問題!
X-Y=6,X的2次+Y的2次=12,X+Y=什麼?A.2B.4C.7 D.18
which is the correct answer? X-Y=6,X的2次+Y的2次=12,X+Y=什麼?
係X的2次+Y的2次 定
X的2次-Y的2次?= =
如果係X的2次-Y的2次
x^2-y^2=12
(x+y)6=12
x+y=2
ans is a
如果係X的2次+Y的2次
咁a,b,c,d都錯 點解x^2-y^2=12
(x+y)6=12
唔明= = {x-y=6 -------------(1)
{x^2+y^2=12----(2)
in (1):x=6+y-------(3)
in (2):x=sqt(12-y^2)----(4)
6+y=sqt(12-y^2)
(6+y)^2=12-y^2
36+12y+y^2=12-y^2
2y^2+12y+24=0
y^2+6y+12=0
之後既step f2未學- - 原帖由 -終場ソ使者- 於 3-11-2008 20:49 發表 http://www.nakuz.com/bbs/images/common/back.gif
{x-y=6 -------------(1)
{x^2+y^2=12----(2)
in (1):x=6+y-------(3)
in (2):x=sqt(12-y^2)----(4)
6+y=sqt(12-y^2)
(6+y)^2=12-y^2
36+12y+y^2=12-y^2
2y^2+12y+24=0
y^2+6y+12=0
之後既step f2未學- -
根本就計唔到~!!
中4學ge方法:
y^2+6y+12=0
△=6^2-4(1)(12)
=-12
=〈0
所以搵唔到y個ans
即a,b,c,d都錯晒 原帖由 ☆雲☆ 於 9-11-2008 20:02 發表 http://www1.nakuz.com/bbs/images/common/back.gif
根本就計唔到~!!
中4學ge方法:
y^2+6y+12=0
△=6^2-4(1)(12)
=-12
=〈0
所以搵唔到y個ans
即a,b,c,d都錯晒
冇錯△<0
兩題跟本就冇ac
但係如果以中二o既角度去念下
a,b,c 呢3個數就算點拆兩減都冇可能係7 只有18有可能- -
ans 應該係d...
p.s.個ans 唔係 y^2+6y+24=0咩- -?? X-Y=6
X^2+Y^2=12
(X-Y)^2=6^2
X^2+Y^2-2XY=36
X^2+Y^2=36+2XY
12=36+2XY
2XY=-24
X+Y=[(X+Y)^2]^(1/2)
X+Y=(X^2+Y^2+2XY)^(1/2)
X+Y=(12-24)^(1/2)
X+Y=(-12)^(1/2)
X+Y=12i!!?(虛數)
我計錯耶- -''!?!?!?
[ 本帖最後由 maxycy 於 9-11-2008 21:51 編輯 ] X-Y=6, x^2+y^2=12, X+Y=?
(X-Y)^2=36
x^2+y^2 - 2xy = 36
12 -2xy = 36
2xy = -24
(X+Y)^2=x^2+y^2+2xy = 12 - 24 = -12
X+Y = 2sqrt3 i
same. 回8#
我諗...
大家都計到12i...
咁睇黎...
真係學2#話...
樓主打錯題目....
題目應是x^2-y^2
而唔係x^2+y^2
[] [] 原帖由 simonluk 於 9-11-2008 20:18 發表 http://www.nakuz.com/bbs/images/common/back.gif
冇錯△<0
兩題跟本就冇ac
但係如果以中二o既角度去念下
a,b,c 呢3個數就算點拆兩減都冇可能係7 只有18有可能- -
ans 應該係d...
p.s.個ans 唔係 y^2+6y+24=0咩- -??
唔係 y^2+6y+24=0 lo...
全式除左2 bo...
24唔洗除??
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