মনে করি,

 (a + b + c) p = (b + c - a) q = (c + a - b) r

      = (a + b - c) s = k

      :. p (a + b + c) = k

      বা, p `= k/(a + b + c)`

      :. `1/p = (a + b + c)/k`........(i)

       আবার,  q (b + c - a) = k

       বা, `q = k/(b + c - a)`

       :. `1/q = (b + c - a)/k`.........(ii)

        আবার, r (c + a - b) = k

        বা,  r = `k/(c + a - b)`

        :. `1/r = (c + a - b)/k`...........(iii)

        এবং  s (a + b - c) = k

        বা,  s =` k/(a + b - c)`

        :.` 1/s = (a + b - c)/k`..........(iv)

    (ii) নং (iii) নং এবং (iv) নং সমীকরণ যোগ করে পাই,

    `1/q + 1/r + 1/s = (b + c - a)/k + (c + a - b)/k + (a + b - c)/k`

    `= 1/k (b + c - a + c + a - b + a + b - c)`

    `= (a + b + c)/k`

    `= 1/p` [(i) নং থেকে]

     :. `1/q + 1/r + 1/s = 1/p`  (প্রমাণিত)

দেওয়া আছে, 

 `p/q = a^2/b^2`

  বা, `(p + q)/(p - q) = (a^2 + b^2)/(a^2 - b^2)` [যোজন-বিয়োজন করে].........(i) 
  
   আবার, `a/b = (sqrt(a + q))/(sqrt(a - q))`

   বা, `a^2/b^2 = (a + q)/(a - q)`; [বর্গ করে] 

   বা, `(a^2 + b^2)/(a^2 - b^2) = (a + q + a - q)/(a + q - a + q)` [যোজন-বিয়োজন করে]

   বা, `(a^2 + b^2)/(a^2 - b^2) = (2a)/(2q)`

    :.` (a^2 + b^2)/(a^2 - b^2) = a/q`..........(ii)

    (i) এবং (ii) নং থেকে পাই,

   `(p + q)/(p - q) = a/q`

    বা, `(p + q)/a = (p - q)/q` (দেখানো হলো)

ক. a, b, c ও d `a/b = b/c = c/d` ক্রমিক সমানুপাতিক হলে

 খ. দেওয়া আছে,` a/b = b/c = c/d`

       ধরি, `a/b = b/c = c/d = k`

          :. c = dk

         b = ck = dk.k = `dk^2`

       `a = bk = dk^2.k = dk^3`

  বামপক্ষ 
     `= (a^3 + b^3)/(b^3 + c^3) = ((dk^3)^3 + (dk^2)^3)/((dk^2)^3 + (dk)^3`)

     `= (d^3k^9 + d^3k^6)/(d^3k^6 + d^3k^3) = (d^3k^3(k^3 + 1))/(d^3k^3(k^3 + 1)) = k^3`

  ডানপক্ষ  
    `= (b^3 + c^3)/(c^3 + d^3) = ((dk^2)^3 + (dk)^3)/((dk)^3 + d^3)`

    `= (d^3k^6 + d^3k^3)/(d^3k^3 + d^3) = (d^3k^3 (k^3 + 1))/(d^3 (k^3 + 1)) = k^3`

     :. `(a^3 + b^3)/(b^3 + c^3) = (b^3 + c^3)/(c^3 + d^3)` (দেখানো হলো)

 
  গ. বামপক্ষ 
      `= a^2b^2 c^2 (1/a^3 + 1/b^3 + 1/c^3)`

      `= (a^2b^2c^2)/a^3 + (a^2b^2c^2)/b^3 + (a^2b^2c^2)/c^3`

      `= (b^2c^2)/a + (a^2c^2)/b + (a^2b^2)/c`

      `= (b^2c^2)/a + (ac)^2/b + (a^2b^2)/c`

      `= (ac.c^2)/a + (b^2)^2/b + (a^2.ac)/c` [:. ac `= b^2`]

      `= c^3 + b^4/b + a^3 = c^3 + b^3 + a^3`

      `= a^3 + b^3 + c^3`

        = ডানপক্ষ

      :. `a^2b^2c^2 (1/a^3 + 1/b^3 + 1/c^3) = a^3 + b^3 + c^3` (প্রমাণিত)

ক. দেওয়া আছে, a, b, c, d ক্রমিক সমানুপাতী।

  সুতরাং `a/b = b/c = c/d`

  ধরি,  `a/b = b/c = c/d = k`

          :. c = dk

  এবং  ` b = ck = dk.k = dk^2`

  এবং  `a = bk = dk^2.k = dk^3`


  খ. ’ক’ থেকে প্রাপ্ত `c = dk, b = dk^2, a = dk^3`

  বামপক্ষ 
    `= (a^2 + b^2 + c^2) (b^2 + c^2 + d^2)`

    `= {(dk^3)^2 + (dk^2)^2 + (dk)^2} {(dk^2)^2 + (dk)^2 + d^2}`

    `= {d^2k^6 + d^2k^4 + d^2k^2} {d^2k^4 + d^2k^2 + d^2}`

   ` = d^2k^2 (k^4 + k^2 + 1) xx d^2 (k^4 + k^2 + 1)`

    `= d^2 xx d^2 k^2 (k^4 + k^2 + 1)^2`

    `= d^4k^2 (k^4 + k^2 + 1)^2`

  ডানপক্ষ  
     `= (ab + bc + cd)^2`

     `= (dk^3 xx dk^2 + dk^2 xx dk + dk xx d)^2`

     `= (d^2k^5 + d^2k^3 + d^2k)^2`

     `= {d^2k (k^4 + k^2 + 1)}^2`

     `= d^4k^2 (k^4 + k^2 + 1)^2`

     :. `(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + sd)^2`(প্রমাণিত)

  গ. দেওয়া আছে, `x = (10ab)/(a + b)`

                  `x/(5a) = (2b)/(a + b)`

                `(x + 5a)/(x - 5a) = (2b + a + b)/(2b - a - b)` [যোজন-বিয়োজন]

                `(x + 5a)/(x - 5a) = (a + 3b)/(b - a)`..........(i)

        আবার, `x = (10ab)/(a + b)`

                 `x/(5b) = (2a)/(a + b)`

                 `(x + 5b)/(x - 5b) = (2a + a + b)/(2a - a - b)` [যোজন-বিয়োজন]

                ` (x + 5b)/(x - 5b) = (3a + b)/(a - b)`............(ii)

                   (i) নং ও (ii) নং যোগ করি,

                  `(x + 5a)/(x - 5a) + (x + 5b)/(x - 5b) = (a + 3b)/(b - a) = (a + 3b - 3a - b)/(b - a)`

                  `= (a + 3b)/(b - a) - (3a + b)/(b - a) = (a + 3b - 3a - b)/(b - a)`

                  `= (2b - 2a)/(b - a) = (2(b - a))/(b - a)`

                    = 2

                  `:. (x + 5a)/(x - 5a) + (x + 5b)/(x - 5b) = 2` (দেখানো হলো)

ক. দেওয়া আছে,

  `(sqrt(1 + x) + sqrt(1 - x))/(sqrt(1 + x - sqrt(1 - x))`  = p..........(i)


    p = 1 হলে (i) থেকে পাই,

   `(sqrt(1 + x) + sqrt(1 - x))/(sqrt(1 + x - sqrt(1 - x)) = 1`

     বা, `sqrt(1 + x) + sqrt(1 - x) = sqrt(1 + x) - sqrt(1 - x)`

     বা,`sqrt(1 + x) + sqrt(1 - x) - sqrt(1 + x) + sqrt(1 - x) = 0`

     বা, `2sqrt(1 - x) = 0`

     বা, `sqrt(1 - x) = 0`

       1 - x = 0 [বর্গ করে] 

        :. x = 1


   খ.


  গ. দেওয়া আছে, 
    `(sqrt(1 + x) + sqrt(1 - x))/(sqrt(1 + x) - sqrt(1 - x)) = p = (2 + sqrt(2))/(2 - sqrt(2))`

      বা, `(sqrt(1 + x) + sqrt(1 - x) + sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x) - sqrt(1 + x) + sqrt(1 - x))`

      বা,`(2 + sqrt(2 + 2) - sqrt(2))/(2 + sqrt(2 - 2) + sqrt(2))` [যোজন-বিয়োজন করে]

      বা,`(2 sqrt(1 + x))/(2 sqrt(1 - x)) = 4/(2 sqrt(2))`

      বা, `(sqrt(1 + x))/(sqrt(1 - x)) = 2/sqrt(2)`

       বা, ` (sqrt(1 + x))/(sqrt(1 - x)) = sqrt(2)`

        বা, `(1 + x)/(1 - x) = 2` [বর্গ করে]

        বা,  1 + x = 2(1 - x)

         বা, 1 + x = 2 - 2x

         বা,  x + 2x = 2 - 1

          বা, 3x = 1

         `:. x = 1/3` (Ans)