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Question Number 214977 by Abdullahrussell last updated on 25/Dec/24

Solve: (((3x−5)^5 −(2x−3)^5 −(x−2)^5 )/((3x−5)^3 −(2x−3)^3 −(x−2)^3 ))=65

Solve:(3x−5)5−(2x−3)5−(x−2)5(3x−5)3−(2x−3)3−(x−2)3=65

Answered by MrGaster last updated on 25/Dec/24

x=4 or x−(5/7)

x=4orx−57

Answered by mehdee7396 last updated on 25/Dec/24

2x−3=a & x−2=b (((a+b)^5 −a^5 −b^5 )/((a+b)^3 −a^3 −b^3 ))=((5a^3 +10a^2 b+10ab^2 +5b^3 )/(3a+3b)) =((5(a^2 +ab+b^2 ))/3)=65 a^2 +ab+b^2 =39 (2x−3)^2 +(2x−3)(x−2)+(x−2)^2 =39 7x^2 −23x−20=0 x=((56)/(14))=4 & x=−(5/7)

2x−3=a&x−2=b(a+b)5−a5−b5(a+b)3−a3−b3=5a3+10a2b+10ab2+5b33a+3b=5(a2+ab+b2)3=65a2+ab+b2=39Missing \left or extra \right7x2−23x−20=0x=5614=4&x=−57

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