Question

\(\sqrt[3]{x+10}\) +\(\sqrt[3]{17-x}\) =3

Answer 1

Với x=17, ta thấy pt TM. Vậy x=17 là nghiệm.

Với , ta có:

\(\Leftrightarrow\sqrt[3]{x+10}-3+\sqrt[3]{17-x}=0\)

\(\Leftrightarrow\dfrac{x-17}{\sqrt[3]{x+10}^2+3\sqrt[3]{x+10}+9}-\dfrac{x-17}{\sqrt[3]{17-x}^2}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{\sqrt[3]{x+10}^2+3\sqrt[3]{x+10}+9}-\dfrac{1}{\sqrt[3]{17-x}^2}\right)=0\)

\(\Rightarrow x=17\left(KTM\right)\). Vì \(\left(\dfrac{1}{\sqrt[3]{x+10}^2+3\sqrt[3]{x+10}+9}-\dfrac{1}{\sqrt[3]{17-x}^2}\right)\ne0\)

Vậy x=17.