If a y - z - x- bz - x - y- cx - y - z- prove that x2a1-bc - y2b 1-ca - z2c1-ab-

From first and second equations we have xa(b+1) = yb(a+1) = z1 ab #160; #160; ....(1)From second and third equations we have x1 bc = yb(c+1) = zc( ...

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Byju's Answer

Standard XII

Mathematics

Different Types of Intervals in Inequality

If a y + z ...

Question

If $a (y + z) = x, b (z + x) = y, c (x + y) = z$ , prove that $\frac{x^{2}}{a (1 - b c)} = \frac{y^{2}}{b (1 - c a)} = \frac{z^{2}}{c (1 - a b)}$ .

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Solution

From first and second equations we have
$\frac{x}{a (b + 1)} = \frac{y}{b (a + 1)} = \frac{z}{1 - a b}$ ....(1)
From second and third equations we have
$\frac{x}{1 - b c} = \frac{y}{b (c + 1)} = \frac{z}{c (b + 1)}$ .....(2)
From first and third equations, we have
$\frac{x}{a (c + 1)} = \frac{y}{1 - a c} = \frac{z}{c (a + 1)}$ .....(3)
From (1) and (2), we have
$\frac{x}{a (b + 1)} \times \frac{x}{1 - b c} = \frac{z}{1 - a b} \times \frac{z}{c (b + 1)}$
$\Rightarrow \frac{x^{2}}{a (1 - b c)} = \frac{z^{2}}{c (1 - a b)}$
From (2) and (3), $\frac{x}{1 - b c} \times \frac{x}{a (c + 1)} = \frac{y}{b (c + 1)} \times \frac{y}{1 - a c}$
$\Rightarrow \frac{x^{2}}{a (1 - b c)} = \frac{y^{2}}{b (1 - a c)}$
Therefore, $\frac{x^{2}}{a (1 - b c)} = \frac{y^{2}}{b (1 - c a)} = \frac{z^{2}}{c (1 - a b)}$

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If a(y+z)=x,b(z+x)=y,c(x+y)=z, prove that x2a(1−bc)=y2b(1−ca)=z2c(1−ab).

If $a (y + z) = x, b (z + x) = y, c (x + y) = z$ , prove that $\frac{x^{2}}{a (1 - b c)} = \frac{y^{2}}{b (1 - c a)} = \frac{z^{2}}{c (1 - a b)}$ .