It is given that x>0, y>0 and √x(√x+√y) = 3√y(√x+5√y).
Find the value of (2x + √xy + 3y) / (x + √xy - y)
Maths(HKMO)
2010-01-01 3:07 am
回答 (2)
2010-01-01 3:28 am
✔ 最佳答案
√x(√x+√y) = 3√y(√x+5√y).x + √(xy) = 3√(xy) + 15y
2√(xy) = x - 15y
(√x)^2 - 2√y√x - 15(√y)^2 = 0
(√x - 5√y)(√x + 3√y) = 0
√x = 5√y or √x = -3√y(rejected)
√x : √y = 5 : 1
So
(2x + √xy + 3y) / (x + √xy - y)
= [2(5^2) + 5*1 + 3(1)^2] / (5^2 + 5*1 - 1^2)
= 58/29
= 2
2010-01-01 3:31 am
From √x (√x+√y)=3√y(√x+5√y),
we can obtain
( x+√xy)=3√xy+15y -------------------(1)
x-2√xy-15y=0----------------------------(2)
(√x+3√y)(√x-5√y)=0, we can obtain
√x=5√y and x=25y
When we sub. √x=5√y and x=25y into the question,
(2x + √xy + 3y) / (x + √xy - y)
=(50y+√25y^2+3y)/(25y+√25y^2-y)
=58y/29y
=2
So, the answer of this question is 2
we can obtain
( x+√xy)=3√xy+15y -------------------(1)
x-2√xy-15y=0----------------------------(2)
(√x+3√y)(√x-5√y)=0, we can obtain
√x=5√y and x=25y
When we sub. √x=5√y and x=25y into the question,
(2x + √xy + 3y) / (x + √xy - y)
=(50y+√25y^2+3y)/(25y+√25y^2-y)
=58y/29y
=2
So, the answer of this question is 2
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091231000051KK01442