我想問 1b) 果題點計...
但唔要: 6^3+2^3
=(6*6*6)+(2*2*2)
=(36*6)+(4*2)
=216+8
= 224
1a)Expand (x+y)(x^2-xy+y^2)
1b)Hence,find the value of 6^3+2^3 without using a calculator.
MATH...好趕架
2007-11-07 1:54 am
回答 (2)
2007-11-07 2:03 am
✔ 最佳答案
你只要做完 1a你便知道
原來...
(x+y)(x^2-xy+y^2) = x^3 + y^3
所以 1b 會用返 1a 條式
6^3 + 2^3
= (6+2)(6^2-6*2+2^2) <- 將 6 和 2 分別代入 x 和 y
= 8*(36-12+4)
= 8*(28)
= 224
記緊
若題目有分題, 且有 hence 或 thus 字樣
便是代表呢個分題同 part a 有關
用返 1a 條式就可以了
參考: Please support me :)
2007-11-07 4:57 am
1a) (x+y)(x2-xy+y2)
=(x)(x2-xy+y2)+(y)(x2-xy+y2)
=(x3-x2y+xy2)+(x2y-xy2+y3)
=x3-x2y+xy2+x2y-xy2+y3
1b) (63+23)
=(6+2)[62-(6)(2)+22]
=(8)(36-12+4)
=(8)(28)
=224
2) 2x-6y/3y-x
= 2x-6y/-(x-3y)
=2(x-3y)/-(x-3y)
=2/-1
=- 2/1
2) -x2+1/x+1
=- (x)(x+1)/x+1
= -x
3) 2p-10r/r2 ÷ 30-15r/6pr
= 2p-10r/r2 ÷ 6pr/3p-15r
= 2(p-5r)/r2 ÷ 6pr/3(p-5r)
= (2)(6p)/(r)(3)
= (2)(2p)/r
= 4p/r
2007-11-06 20:57:54 補充：
Question 2= -2/1should be:= -2
=(x)(x2-xy+y2)+(y)(x2-xy+y2)
=(x3-x2y+xy2)+(x2y-xy2+y3)
=x3-x2y+xy2+x2y-xy2+y3
1b) (63+23)
=(6+2)[62-(6)(2)+22]
=(8)(36-12+4)
=(8)(28)
=224
2) 2x-6y/3y-x
= 2x-6y/-(x-3y)
=2(x-3y)/-(x-3y)
=2/-1
=- 2/1
2) -x2+1/x+1
=- (x)(x+1)/x+1
= -x
3) 2p-10r/r2 ÷ 30-15r/6pr
= 2p-10r/r2 ÷ 6pr/3p-15r
= 2(p-5r)/r2 ÷ 6pr/3(p-5r)
= (2)(6p)/(r)(3)
= (2)(2p)/r
= 4p/r
2007-11-06 20:57:54 補充：
Question 2= -2/1should be:= -2
參考: me
收錄日期: 2021-04-13 14:22:13
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