If(b+c-a)x =(c+a-b)y =(a+b-c)z=2 then find the value of(1/y+1/z)(1/z+1/x) (1/x+1/y)
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(b+c-a)x=(c+a-b)y=(a+b-c)z=2
∴, (b+c-a)x=2
or, x=2/(b+c-a)
or, 1/x=(b+c-a)/2
Similarly, 1/y=(c+a-b)/2 and 1/z=(a+b-c)/2
∴, (1/y+1/z)(1/z+1/x)(1/x+1/y)
={(c+a-b)/2+(a+b-c)/2}{(a+b-c)/2+(b+c-a)/2}{(b+c-a)/2+(c+a-b)/2}
={(c+a-b+a+b-c)/2}{(a+b-c+b+c-a)/2}{(b+c-a+c+a-b)/2}
=(2a/2)(2b/2)(2c/2)
=abc
∴, (b+c-a)x=2
or, x=2/(b+c-a)
or, 1/x=(b+c-a)/2
Similarly, 1/y=(c+a-b)/2 and 1/z=(a+b-c)/2
∴, (1/y+1/z)(1/z+1/x)(1/x+1/y)
={(c+a-b)/2+(a+b-c)/2}{(a+b-c)/2+(b+c-a)/2}{(b+c-a)/2+(c+a-b)/2}
={(c+a-b+a+b-c)/2}{(a+b-c+b+c-a)/2}{(b+c-a+c+a-b)/2}
=(2a/2)(2b/2)(2c/2)
=abc
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