Đáp án: `S={5;6;7}`
Giải thích các bước giải:
$ĐKXĐ:x∈R$
Đặt $\sqrt[3]{7-x}=a;\sqrt[3]{x-5}=b$
$⇒a^3=7-x;b^3=x-5$
$⇒a^3-b^3=7-x-x+5=12-2x=2(6-x)$
Phương trình đã cho trở thành:
`\frac{a-b}{a+b}=\frac{1}{2}(a^3-b^3)`
$⇒2(a-b)=(a^3-b^3)(a+b)=(a+b)(a^2+ab+b^2)(a-b)$
$⇔(a-b)[(a+b)(a^2+ab+b^2)-2]=0$
$⇔\left[ \begin{array}{l}a-b=0\\(a+b)(a^2+ab+b^2)-2=0\end{array} \right.$
-Trường hợp 1: Nếu $a=b⇔\sqrt[3]{7-x}=\sqrt[3]{x-5}$
$⇔7-x=x-5⇔x+x=7+5⇔2x=12⇔x=6$
-Trường hợp 2: Nếu $(a+b)(a^2+ab+b^2)-2=0$
$⇔a^3+2a^2b+2ab^2+b^3=2$
$⇔(7-x)+(x-5)+2.(\sqrt[3]{7-x})^2.\sqrt[3]{x-5}+2..\sqrt[3]{7-x}.(\sqrt[3]{x-5})^2=2$
$⇔2.\sqrt[3]{(7-x)^2(x-5)}+2\sqrt[3]{(7-x)(x-5)^2}=0$
$⇔2.\sqrt[3]{(7-x)^2(x-5)}=-2\sqrt[3]{(7-x)(x-5)^2}$
$⇔8(7-x)^2(x-5)=-8(7-x)(x-5)^2$
$⇔(7-x)^2(x-5)=(x-7)(x-5)^2$
$⇔(x-7)^2(x-5)-(x-7)(x-5)^2=0$
$⇔(x-7)(x-5)(x-7-x+5)=0$
$⇔(x-7)(x-5)(-2)=0⇔(x-7)(x-5)=0$
$⇔\left[ \begin{array}{l}x-7=0\\x-5=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=7\\x=5\end{array} \right.$
