Summer Clouds moderators
30/06/2017 at 09:00-
Trần Nhật Dương 30/06/2017 at 14:34We have: x.y=14x.y=14 and y.z=10y.z=10 and z.x=35z.x=35
<=>x.y.y.z.z.x=14.10.35x.y.y.z.z.x=14.10.35
<=>x2.y2.z2=4900x2.y2.z2=4900
<=>(x.y.z)2=702(x.y.z)2=702
=> x.y.z=70x.y.z=70 <1>
From the context and <1>
=> x=(x.y.z):(y.z)=70:10=7x=(x.y.z):(y.z)=70:10=7
=> y=(x.y):x=14:7=2y=(x.y):x=14:7=2
=> z=(z.x):x=35:7=5z=(z.x):x=35:7=5
So x=7;y=2;z=5
=>The value of x+y+z is 7+2+5=14
So the answer of the problem is C.14 .
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Tran Nhat Duong, do not copy my answer !!!
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Khoa Nguyễn Văn 05/07/2017 at 11:20Ta có:\(x\times y\times y\times z\times x\times z=x^2\times y^2\times z^2=14\times10\times35=4900=\left(x\times y\times z\right)^2=70^2\)
\(\Rightarrow x\times y\times z=70\)
\(\Rightarrow x=7;y=2;z=5\)
\(\Rightarrow x+y+z=7+2+5=14\)
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(CONTINUE)
=> The value of \(x+y+z=7+2+5=14\)
So the answer is C: 14
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Lê Quốc Trần Anh,you forgot to find the sum of x,y,z
The answer is : 7 + 2 + 5 = 14 (C)
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We have: \(x.y=14\) and \(y.z=10\) and \(z.x=35\)
<=>\(x.y.y.z.z.x=14.10.35\)
<=>\(x^2.y^2.z^2=4900\)
<=>\(\left(x.y.z\right)^2=70^2\)
=> \(x.y.z=70\) <1>
From the context and <1>
=> \(x=\left(x.y.z\right):\left(y.z\right)=70:10=7\)
=> \(y=\left(x.y\right):x=14:7=2\)
=> \(z=\left(z.x\right):x=35:7=5\)
So \(x=7;y=2;z=5\)