lok yin 發問於 科學及數學數學 · 1 十年前

[[[幫手唔識做---identity+change of subject

有d要用identity

(5a+b)^2 -(5a-b)^2

(3x+y)^2 -(2x+3y)^2

4x^2 -4x(2y-z) +(2y-z)^2

(3x+y)^2 -2(3x+y)(x-y)+ (x-y)^2

(x+y)^2 -2(x+y) +1-x^2

(a+b)^2 -2(a+b)(a-b) -(a-b)^2

(x+y)^2 -4(x^2-y^2) +4(x-y)^2

--------------------------

change of subject

mgh = E - (1/2)mv^2

2mgh = 2E - mv^2呢個step點來?

h = (2E - mv^2)/(2mg)

仲有E-(1/2mv^2)乘1/mg點計

更新:

I am so sorry that I didnt tell you those questions are using identities to expand and simplify but not factorize.

Sorry again for my carelessness

1 個解答

評分
  • Rinka
    Lv 5
    1 十年前
    最愛解答

    1st identity : a^2-b^2 = (a+b) (a-b)

    比如 5^2 - 4^2 = (5+4) (5-4) = 9

    所以第一題

    (5a+b)^2 -(5a-b)^2

    = [(5a+b) + (5a-b)] [(5a+b)-(5a-b)]

    = (5a + b + 5a -b) (5a + b - 5a +b)

    = 10a (2b)

    = 20ab

    第二題都係一樣 , 試試自己做

    答案 = (x-2y) (5x+4y)

    2nd identity = (a-b)^2 = a^2 -2ab +b^2

    比如 (2x-3y)^2 = 4x^2 - 6xy + 9y^2

    所以第三題

    4x^2 -4x(2y-z) +(2y-z)^2

    = (2x)^2 - 2(2x)(2y-z) + (2y-z)^2 (把題目造成 某某2次 - 2 (某某) (某) + (某)2次 的形式)

    = [(2x) - (2y-z)]^2

    = (2x-2y+z)^2

    第四 , 六 , 七題一樣 , 自己做做看

    第四題答案 = (2x-2y)^2 = 4(x-y)^2

    第六題答案 = 4b^2

    第七題答案 = (3y-x)^2

    第五題要用哂1st identity 同 2nd identity

    (x+y)^2 -2(x+y) +1-x^2

    = [(x+y) -1] ^2 - x^2

    = (x+y-1)^2 - x^2

    = (x+y-1-x)(x+y-1+x)

    = (y-1)(2x+y-1)

    ____________________________________________________

    change of subject

    個step係兩面一齊乘以2

    mgh x 2 = 2mgh

    E x 2 = 2E

    (1/2)mv^2 x 2 = mv^2

    條題目應該係 有個 [h] o係最後

    即係寫到左面得一個h , 右面冇哂h就叫做完

    h = (2E - mv^2)/(2mg) 就係答案了 , 唔駛計

    有講漏咩既就再問啦?

    2007-01-01 17:29:14 補充:

    expand 的話只係做多一步 , 就係將d 括號乘番問1 , 20ab2 , 5x^2 -6xy -8y^23 , 4x^2 4y^2 z^2 4xz -4yz4 , 4x^2 -8xy 4y^25 , y^2 2xy -2x -2y 16 , 4b^27 , 9y^2 -6xy x^2(如果真係expand and simplify 就係咁啦)

還有問題嗎?立即提問即可得到解答。