If (7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d);
prove that a : b = c : d.
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Given,
(7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d)
∴7a+8b7a−8b=7c+8d7c−8d\therefore \dfrac{7a + 8b}{7a - 8b} = \dfrac{7c + 8d}{7c - 8d}∴7a−8b7a+8b=7c−8d7c+8d
Applying componendo and dividendo: ⇒7a+8b+7a−8b7a+8b−(7a−8b)=7c+8d+7c−8d7c+8d−(7c−8d)⇒14a16b=14c16d⇒ab=cd.\Rightarrow \dfrac{7a + 8b + 7a - 8b}{7a + 8b - (7a - 8b)} = \dfrac{7c + 8d + 7c - 8d}{7c + 8d - (7c - 8d)} \\[1em] \Rightarrow \dfrac{14a}{16b} = \dfrac{14c}{16d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.⇒7a+8b−(7a−8b)7a+8b+7a−8b=7c+8d−(7c−8d)7c+8d+7c−8d⇒16b14a=16d14c⇒ba=dc.
Hence, proved that a : b = c : d.
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If a : b = c : d, prove that :
5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d
xa + yb : xc + yd = b : d
If x = 6aba+b\dfrac{6ab}{a + b}a+b6ab, find the value of :
x+3ax−3a+x+3bx−3b\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}x−3ax+3a+x−3bx+3b.
If a = 462+3\dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}2+346, find the value of :
a+22a−22+a+23a−23\dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}}a−22a+22+a−23a+23.
