(i) দেওয়া আছে, a : b = b : c,
বা, \[ \frac{a}{b} = \frac{b}{c} \]
\[\therefore b^2 = ac \]
বামপক্ষ = \[ \frac{a}{c} \]
ডানপক্ষ = \[ \frac{a^2 + b^2}{b^2 + c^2} \]
= \[ \frac{a^2 + ac}{ac + c^2} \]
= \[ \frac{a (a + c)}{c (a + c)} \]
= \[ \frac{a}{c} \]
অর্থাৎ, \[ \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2} \]
Class 9-10 math ex 11.1 solution
(ii) দেওয়া আছে, a : b = b : c,
বা, \[ \frac{a}{b} = \frac{b}{c} \]
\[\therefore b^2 = ac \]
বামপক্ষ = \[a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3}) \]
= \[ \frac{a^2b^2c^2}{a^3} + \frac{a^2b^2c^2}{b^3} + \frac{a^2b^2c^2}{c^3} \]
= \[ \frac{b^2c^2}{a} + \frac{a^2c^2}{b} + \frac{a^2b^2}{c} \]
= \[ \frac{ac.c^2}{a} + \frac{(b^2)^2}{b} + \frac{a^2.ac}{c} \] [\Since b^2 = ac]
= \[ \frac{a.c^3}{a} + \frac{b^4}{b} + \frac{a^3c}{c} \]
= \[ c^3 + b^3 + a^3 \]
= \[ a^3 + b^3 + c^3 \]
= ডানপক্ষ
অর্থাৎ, \[a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3}) = a^3 + b^3 + c^3 \]
(iii) দেওয়া আছে, a : b = b : c,
বা, \[ \frac{a}{b} = \frac{b}{c} \]
\[\therefore b^2 = ac \]
বামপক্ষ = \[ \frac{abc (a + b + c)^3} {(ab + bc + ca)^3} \]
=\[ \frac{b. b^2 (a + b + c)^3}{(ab + bc + b^2)^3} \] \[ [\because b^2 = ac]\]
= \[ \frac{b^3 (a + b + c)^3}{{b (a + c + b)}^3} \]
= \[\frac{b^3 (a + b + c)^3}{b^3 (a + b + c)^3} \]
= 1 = ডানপক্ষ
(iv) দেওয়া আছে, a : b = b : c,
বা, \[ \frac{a}{b} = \frac{b}{c} \]
\[\therefore b^2 = ac \]
১ম পক্ষ = a – 2b + c
২য় পক্ষ = \[ \frac{(a – b)^2}a \]
= \[ \frac{a^2 – 2ab + b^2}{a} \]
= \[ \frac{a^2 – 2ab + ac}{a} [\since b^2 = ac] \]
= \[ \frac{a(a – 2b + c)}{a} \]
= a – 2b + c
৩য় পক্ষ = \[ \frac{(b – c)^2}{c} \]
= \[ \frac{b^2 – 2bc + c^2}{c} \]
= \[ \frac{ac – 2bc + c^2}{c} [\since b^2 = ac] \]
= \[ \frac{c(a – 2b + c)}{c} \]
= a – 2b + c
অর্থাৎ, a – 2b + c = \[ \frac{(a – b)^2}a \] = \[ \frac{(b – c)^2}{c} \]
