Partial fraction (Aanshik bhinnh) shortcuts tricks part-1 in Hindi

Hello, friends! aapka ek baar phir mai "Maths Tricks in hindi " swagat karta hu. Ek baar phir aapk liye mathematics ka ek new tricks lekar aaya hu.
Also read-:Multiply tricks of two digits
Aaj mai jis mathematics k tricks k baare me batauga uska naam hai, "Partial fraction " . Jisse hindi me "Aanshik bhinnh"(आंशिक  भिन्न ) kahate hai. To aaj hum sikkhege ki -:

Partial fraction ( Aanshik bhinnh) k question ko tricks se kaise solve karte hai  ? Part-1

Generally ,mathematics  me bhot jagah par Aanshik bhinnh ka use hai. Aanshik bhinnh(Partial fraction ) ka vishes use Integral functions  ko solve krne me karte hai. Aayiye general method se partial fraction ka ek question solve karte hai-:
General method -:
Ques-:          2x + 3                                        
                   (x+1)(x-3) 
Upper diye   question ko "partial fraction " me separate karna hai? 
Solution -:      question Diya hai-:
                         
                             2x+3  
                        (x+1)(x-3)
Isse equation (1) Maan lete hai.
Ab equation (1) ko separate karte -:
                    2x+3     =     A     +    B   
               (x+1)(x-3)      (x+1)    (x-3)
Isse equation (2) maan lete hai. Equation (2) me "A" and "B" ki value find karna hai. To equation (2) ko solve karne pr -:
2x+3=A(x-3)+B(x+1)............(3)
Ab "A" aor "B" ki value find karege -:
   *  Jab (x+1)=0
      tab        x=-1 equation (3) me put karne par -:
    -2+3=A(-1-3)
  1=-4A ,    A= -1/4
*  jab (x-3)=0 tab x=3 again equation (3) me put karne par -:
2(3)+3=B(3+1)
9=4B, B =9/4
A and B ki value equation (2) me put karne par exact " partial fraction " hai-:
        -1       +      9     
     4(x+1)      4(x-3)  
Answer. 


Ye to tha general method ab issi question ko tricks se karte hai-:
Tricks method-:

"A" ki value find karne ka tricks -:

1. "A" ki value nikalane k liye equation (2) k Right hand side (R.H.S.) ke "A" k Har (denomenoter) ki value zero rakhege. 

 Tab x+1=0, x=-1 ab x=-1 ko equation (1) me rakhege wahi "A" ki value hogi. Lekin equation (1) k denometor se (x+1) wale part ko remove kar dege. Jaisa Maine niche kiya hair-:

  A=         2x+3   

                (x-3)

A=[ 2(-1)+3]/[-1-3]=-1/4

"B" ki value find karne ka tricks -:
Phir se same "B" k liye karenge. Equation (2) k R.H.S. se "B" k denomenoter (x-3)=0 ,x=3 ko equation (1) me rakhege. Lekin equation (1) k denometor se (x-3) ko emove ker denge-:
B =  (2x+3)/(x+1) 
    = [2(3)+3]/(3+1)
    = 9/4 
Bus ab "A" aor "B" ki value equation(2) me put(rakh) denge
    -1    +    9   
4(x+1)    4(x-3)
 Answer. 
To dekha aapne tricks se kitini aashani se partial fraction solve hogaya. Aayie kuch aor example dekhte hai -:
Example -1.              2x+3   
                               (x+1)(x+2)
Tricks-:    Step-1.      qusetion diya hai -:
                         2x+3  
                     (x+1)(x+2)    
Isse equation (1) maan lete hai. 
Step-2.           2x+3    =      A     +     B    
                    (x+1)(x+2)    (x+1)      (x+2)
Isse equation (2) maante hai. 
Step-3.  "A" ki value k liye -:
              * Equation (2) k R.H.S. me "A" k denomenetor ko zero k equal krege. 
              x+1=0, x=-1
              * x=-1 equation (1) me rakhege. But (x+1) ko remove karne k baad. 
         "A" = (2x+3)/(x+2)
          A = [2(-1)+3]/(-1+2)
          A= 1
Step-3.  "B" ki value k liye -:
             *  Equation (2) k R.H.S me "B" k denomenetor ko -:
                 x+2=0, x=-2
             * x=-2 equation(1) me rakege. But iss baar (x+2) ko remove krege. 
B=  (2x+3)/(x+1)
   = [2(-2)+3]/(-2+1)
B= 1
Step-4.  "A" and "B" ki value equation (2) me rakhne par-:
          2x+3   =    1     +     1   
    (x+1)(x+2)    (x+1)     (x+2)
Ans.
Example-2.       3x+13   
                        (x+1)(x+2)
Tricks-:     3x+13   =   A   +   B    
               (x+1)(x+2)   (x+1)   (x+2)
"A" k liye (x+1)=0 , x=-1
A = (3x+13)/(x+2) issme x=-1 rakhe-:
A= [3(-1)+13]/(-1+2)
A= 10

"B" k liye (x+2)=0, x=-2
B=(3x+13)/(x+1)
B=[3(-2)+13]/(-2+1)
B=7/(-1)=-7
Tab "A" aor "B" ki value rakhne par-:
    3x+13     =    10   -    7   
(x+1)(x+2)     (x+1)    (x+2)
Ans.
Example -3.      2x+1   
                       (x^2-5x+6)  
Upper diye question me "x^2= square of x " hai. Ya isse hindi me "x square " kahate hai. 
Tricks -: sabse pahale iss question me (x^2-5x+6) ka gudankhand( factor ) karna hoga.
Gudankhand karne par -:
x^2-5x+6= (x-2)(x-3)

      2x+1    =      2x+1   
(x^2-5x+6)     (x-2)(x-3)

    2x+1      =     A   +   B   
(x-2)(x-3)      (x-2)     (x-3)
Uppar wale ko equation (1) maante hai ,aor "A" aor "B" ki value nikalte hai.
"A" k liye (x-2)=0,x=2
"A"=(2x+1)/(x-3)  issme x=-2 rakhe-:
A= [2(2)+1]/(2-3)
A=-5
"B" k liye (x-3)=0,x=3
B=(2x+1)/(x-2)
B=[2(3)+1]/(3-2)
B=7
    2x+1     =     -5   +    7   
x^2-5x+6      (x-2)     (x-3)
Ans.
Example -4.      3x^2+12x+11  
                        (x+1)(x+2)(x+3)
Ko partial fraction me change karna hai ?
Tricks-: step-1. 
          3x^2+12x+11   =   A   +   B   +   C  
         (x+1)(x+2)(x+3)   (x+1) (x+2) (x+3)
Isse equation (1) maan lete hai.
Step-2.
   (i)     "A" k liye (x+1)=0,x=-1
Tab "A"=[3x^2+12x+11]/[(x+2)(x+3)]
      x= -1 rakhane par-:
"A"=[3(-1)^2+12(-1)+11]/[(-1+2)(-1+3)]
"A"=2/2=1
(ii) "B" k liye (x+2)=0,x=-2
"B"=[3x^2+12x+11]/[(x+1)(x+3)]
x=-2 rakhne par -:
"B"=[3(-2)^2+12(-2)+11]/(-2+1)(-2+3)
"B"=-1/-1
"B"=1
(iii) "C" k liye (x+3)=0,x=-3
"C"=[3x^2+12x+11]/(x+1)(x+2)
x=-3 rakhne par-:
"C"=[3(-3)^2+12(-3)+11]/(-3+1)(-3+2)
"C"=2/2
"C"=1
Step-3. "A","B" aor "C" ki value equation (1) me rakhne par-:
  3x^2+12x+11  =    1   +    1   +   1   
(x+1)(x+2)(x+3)   (x+1)  (x+2)  (x+3)
Answer.

To dekha friends tricks se partial fraction kitani aasani aor jaldi se solve ho jata hai. Ye post abhi khatam nhi huaa. Isska second part abhi baki hai. Jisse Mai jald hi post karuga.

Friends aaj k liye itana hi, milate hai agale post. Aaj ka post aapko kaisa laga?
Ye batana mat bhuliyega. Agar apka koi sujhav ya Sikayat hai to hame jarur bataye. Thank you !