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kamlesh verma

· started a discussion

· 1 Months ago

c (Question: 207726.0 )

sollution

Question:
If \(\cfrac{x^2}{by +cz}\) = \(\cfrac{y^2}{cz + ax}\) = \(\cfrac{z^2}{ax + by}\) = 2, then  \(\cfrac{c}{2c + z}\)  + \(\cfrac{b}{2b + y}\)  + \(\cfrac{a}{2a + x}\)equals to –
Options:
A) 3
B) 2
C) 4
D) \(\cfrac{1}{2}\)
Solution:
Ans: (d)


= \(\cfrac{1}{2}\)  \(\left [ \cfrac {cz}{ax + by + cz} \ + \cfrac{by}{ax + by + cz} + \cfrac{ax}{ax + by + cz} \ \ \right ]\) 

= \(\cfrac{1}{2}\) = \(\left [ \cfrac {ax + by + cz}{ax + by + cz} \ \ \ \ \right ]\) = \(\cfrac{1}{2}\)

Knowledge Expert

· commented

· 1 Months ago

Dear Student,

Explanation-
c/(2c+z) + b/(2b+y)+c/(2c+z)
according to given condition
we find the value of
x^2= 2(by+cz) , y^2= 2(cz+ax), z^2= 2(ax+by)
We have the value of x^2, y^2, z^2, thus we create this term by multiplying X,Y and z

Z*c/ z*(2c+z) + y* b/ y* (2b+y) + x*a/ x*(2a+x)
now the equation is,

Z*c/ (2cz+z^2) + y* b/ (2by+y^2) + x*a/ (2ax+x^2)

Putting the value of x^2, y^2, z^2

after that simplification of equation, we get the valve.

please read solution and try again.

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