1.))) 4 ( 2q + 7 )^2 - 25 ( q - 6 )^2 = 0
# 要用factor method的。
要個ans係 eg:: x^2 - 4x + 4
(x-1) (x-3) = 0
then,x=1 or x=3
咁樣果d....
呢條的ans. 係 x=44 or x= 16/9
想問點做??.....(要step....)
唔該曬....
中4 math 一條
2007-09-15 3:20 am
回答 (5)
2007-09-15 3:27 am
✔ 最佳答案
4 ( 2q + 7 )^2 - 25 ( q - 6 )^2 = 0[2 (2q+7) + 5 (q-6)] [2( 2q+7)-5 (q-6)]=0
(4q+14+5q-30) (4q+14-5q+30)=0
(9q-16) (-q+44)=0
q =16/9 or 44
2007-09-15 9:04 am
4 ( 2q + 7 )^2 - 25 ( q - 6 )^2 = 0
2^2 (2q + 7)^2 - 5^2 (q - 6 )^2=0
[2 (2q + 7)]^2 - [ 5 (q - 6)]^2=0
(4q + 14)^2 - (5q - 30)^2 =0 (使用平方差公式a^2 - b^2=(a+b)(a-b)
[4q + 14 +( 5q - 30)]*[(4q + 14 - (5q - 30)]=0
(9q - 16)(44 - q)=0
ANS:q=16/9 or q=44
2^2 (2q + 7)^2 - 5^2 (q - 6 )^2=0
[2 (2q + 7)]^2 - [ 5 (q - 6)]^2=0
(4q + 14)^2 - (5q - 30)^2 =0 (使用平方差公式a^2 - b^2=(a+b)(a-b)
[4q + 14 +( 5q - 30)]*[(4q + 14 - (5q - 30)]=0
(9q - 16)(44 - q)=0
ANS:q=16/9 or q=44
2007-09-15 3:39 am
4 ( 2q + 7 )^2 - 25 ( q - 6 )^2 = 0
4(4q^2+28q+49)-25(q^2-12q+36)=0
16q^2+112q+196-25q^2+300q-900=0
-9q^2+412q-704=0
(9q-16)(q-44)=0
q=44 or q= 16/9
4(4q^2+28q+49)-25(q^2-12q+36)=0
16q^2+112q+196-25q^2+300q-900=0
-9q^2+412q-704=0
(9q-16)(q-44)=0
q=44 or q= 16/9
2007-09-15 3:30 am
4(4q^2+28q+49)-25(q^2-12q+36)=0
-9q^2+412q-704=0
9q^2-412q+704=0
(9q-16)(q-44)=0
q=44 or q=16/9
-9q^2+412q-704=0
9q^2-412q+704=0
(9q-16)(q-44)=0
q=44 or q=16/9
參考: by自己
2007-09-15 3:28 am
4(2q + 7)2 – 25(q - 6)2 = 0
4(4q2 + 28q + 49) – 25(q2 – 12q + 36) = 0
-9q2 + 412q – 704 = 0
9q2 – 412q + 704 = 0
(q - 44)(9q - 16) = 0
q = 16/9 or 44
4(4q2 + 28q + 49) – 25(q2 – 12q + 36) = 0
-9q2 + 412q – 704 = 0
9q2 – 412q + 704 = 0
(q - 44)(9q - 16) = 0
q = 16/9 or 44
參考: Myself~~~
收錄日期: 2021-04-13 18:34:51
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