b: (3x-2)^5+(5-x)^5+(-2x-3)^5=0
Đặt a=3x-2; b=-2x-3
Pt sẽ trở thành:
a^5+b^5-(a+b)^5=0
=>a^5+b^5-(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)=0
=>-5a^4b-10a^3b^2-10a^2b^3-5ab^4=0
=>-5a^4b-5ab^4-10a^3b^2-10a^2b^3=0
=>-5ab(a^3+b^3)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2+2ab)=0
=>-5ab(a+b)(a^2+b^2+ab)=0
=>ab(a+b)=0
=>(3x-2)(-2x-3)(5-x)=0
=>\(x\in\left\{\dfrac{2}{3};-\dfrac{3}{2};5\right\}\)