Bài 26 trang 78 sách bài tập Đại số 10

Đkxđ: \(\dfrac{1}{2}-x\le0\) \(\Leftrightarrow x\le\dfrac{1}{2}\).
Đặt: \(a=\sqrt[3]{\dfrac{1}{2}x};b=\sqrt{\dfrac{1}{2}-x}\) \(\left(a,b\in R;b>0\right)\).
Khi đó ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+b=1\\2a^3+b=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\4a^3+2\left(1-a\right)=1\end{matrix}\right.\)
\(\Rightarrow4a^3+2\left(1-a\right)=1\)\(\Leftrightarrow4a^3+2a^2-4a+1=0\)
\(\Leftrightarrow\left(4a^3-a\right)+\left(2a^2-3a+1\right)=0\)\(\Leftrightarrow a\left(2a-1\right)\left(2a+1\right)+\left(2a-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left(2a-1\right)\left[a\left(2a+1\right)+a-1\right]=0\)
\(\Leftrightarrow\left(2a-1\right)\left(2a^2+2a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a-1=0\\2a^2+2a-1=0\end{matrix}\right.\)\(\left[{}\begin{matrix}a=\dfrac{1}{2}\\a=\dfrac{-1+\sqrt{3}}{2}\\a=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\).
Với \(a=\dfrac{1}{2}\) \(\Leftrightarrow\sqrt[3]{\dfrac{1}{2}x}=\dfrac{1}{2}\)\(\Leftrightarrow x=\left(\dfrac{1}{2}\right)^3:\dfrac{1}{2}=\dfrac{1}{4}\) (tm)
Với \(a=\dfrac{-1+\sqrt{3}}{2}\)\(\Leftrightarrow\sqrt[3]{\dfrac{1}{2}x}=\dfrac{-1+\sqrt{3}}{2}\)\(\Leftrightarrow x=\left(\dfrac{-1+\sqrt{3}}{2}\right)^3:\dfrac{1}{2}=\dfrac{\left(-1+\sqrt{3}\right)^3}{4}\) (tm).
Với \(a=\dfrac{-1-\sqrt{3}}{2}\Rightarrow x=\dfrac{\left(-1-\sqrt{3}\right)^3}{4}\) (tm).
Vậy phương trình có ba nghiệm là:\(\dfrac{1}{4};\dfrac{\left(-1+\sqrt{3}\right)^3}{4};\dfrac{\left(-1-\sqrt{3}\right)^3}{4}\).