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Question Number 51153 by Gulay last updated on 24/Dec/18

Commented by Gulay last updated on 24/Dec/18

Sir could you help me?

Sircouldyouhelpme?

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

3)(((79+21)(79−21))/(58))−(((83−17)(83^2 +83×17+17^2 ))/(17^2 +17×83+83^2 )) =((100×58)/(58))−66 =100−66 =34

3)(79+21)(79−21)58−(83−17)(832+83×17+172)172+17×83+832=100×5858−66=100−66=34

Commented by Gulay last updated on 24/Dec/18

thank you so much sir

thankyousomuchsir

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

1)77=a 23=b 75=c 25=d 10^2 =c+d ((a^3 −b^3 )/(a^2 +ab+b^2 ))+((c^2 −d^2 )/(c+d)) (((a−b)(a^2 +ab+b^2 ))/(a^2 +ab+b^2 ))+(((c+d)(c−d))/(c+d)) a−b+c−d 77−23+75−25 =54+50 =104

1)77=a23=b75=c25=d102=c+da3−b3a2+ab+b2+c2−d2c+d(a−b)(a2+ab+b2)a2+ab+b2+(c+d)(c−d)c+da−b+c−d77−23+75−25=54+50=104

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

2)((a^3 +b^3 )/(b^2 −ab+a^2 ))−(((75+25)(75−25))/(2500)) (a+b)−((100×50)/(2500)) 77+23−2=98

2)a3+b3b2−ab+a2−(75+25)(75−25)2500(a+b)−100×50250077+23−2=98

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