If x-y-z-a-y-z-x-b-z-x-y-c- then 1-1-a-1-1-b-1-1-c is equal to

The correct option is C 2 x/y+z=a⇒ 1+x/y+z=a+1 ∴1/1+a=y+z/x+y+z etc ∴ GE=1/x+y+z [ y+z+x+x+y ] =2 ( x+y+z )/ ( x+y+z ) =2 ...

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Elimination Method of Finding Solution of a Pair of Linear Equations

If x/y+z=a;...

Question

If $\frac{x}{y + z} = a; \frac{y}{z + x} = b; \frac{z}{x + y} = c;$ then $\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c}$ is equal to

a + b + c

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Solution

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Similar questions

a = \frac{x}{y + z}

b = \frac{y}{z + x}

c = \frac{z}{x + y}

\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} = 1

x = l o g_{a} (b c), y = l o g_{b} (c a), z = l o g_{c} (a b),

\frac{log x}{b - c} = \frac{log y}{c - a} = \frac{log z}{a - b}

a, b, c > 0

x, y, z ϵ R

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a^{x} + a^{- x})^{2} & (a^{x} - a^{- x})^{2} & 1 (b^{y} + b^{- y})^{2} & (b^{y} - b^{- y})^{2} & 1 (c^{z} + c^{- z})^{2} & (c^{z} - c^{- z})^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

a, b, c > 0

x, y, z \in R

∣ ∣ ∣ ∣ ∣ \begin{matrix} {(a^{x} + a^{- x})}^{2} & {(a^{x} - a^{- x})}^{2} & 1 {(b^{y} + b^{- y})}^{2} & {(b^{y} - b^{- y})}^{2} & 1 {(c^{z} + c^{- z})}^{2} & {(c^{z} - c^{- z})}^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

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