Q.1
^3開方(x^5)=32
Q.2
開方{[8^(2k-4)] / [16^(k-1)]}
f.4 math
2006-11-07 5:00 am
回答 (4)
2006-11-07 5:25 am
✔ 最佳答案
Q.1 你的式是不是如下所示√(x5)=32
(x5/2)=25
x1/2 = 2
x = 4
Q.2
√{[8(2k-4)] / [16(k-1)]}
= √{[23(2k-4)] / [24(k-1)]}
= √{[23(2k-4)-4(k-1)]}
= √{[26k-12-4k+4]}
= √{[22k-8]}
= 2k-4
2006-11-07 5:43 am
Q.1
我諗你Q.1係問x^5既開方3次=32
so:x^5既開方3次=32
x^5/3=32
(x^5/3)^3/5=32^3/5
x=8
Q.2
開方{[8^(2k-4)] / [16^(k-1)]}
開方{[2^(6k-12)] / [2^(4k-4)]}
開方[(2^2k)/(2^8)]
(2^k)/16
我諗你Q.1係問x^5既開方3次=32
so:x^5既開方3次=32
x^5/3=32
(x^5/3)^3/5=32^3/5
x=8
Q.2
開方{[8^(2k-4)] / [16^(k-1)]}
開方{[2^(6k-12)] / [2^(4k-4)]}
開方[(2^2k)/(2^8)]
(2^k)/16
2006-11-07 5:32 am
Q.1
X^5 = 32^ 3
X = 5 √ 32^ 3
X = 8
Q.2
=√ 8^(2k-4) / 16^(k-1)
=√2^3(2k-4) / 2^4(k-1)
=√2^(6k-12) / 2^ (4k-4)
=√2^(6k-12) - (4k-4)
=2^1/2(2k-8)
=2^(k-4)
X^5 = 32^ 3
X = 5 √ 32^ 3
X = 8
Q.2
=√ 8^(2k-4) / 16^(k-1)
=√2^3(2k-4) / 2^4(k-1)
=√2^(6k-12) / 2^ (4k-4)
=√2^(6k-12) - (4k-4)
=2^1/2(2k-8)
=2^(k-4)
2006-11-07 5:11 am
第一題唔明
q2
2^k/16
q2
2^k/16
收錄日期: 2021-04-20 13:14:19
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