Bài 15:
\(x^3-x+1=\sqrt[3]{2x-1}\left(1\right)\)
Đặt \(u=\sqrt[3]{2x-1}\Rightarrow u^3=2x-1\). Ta có hệ phương trình:
\(\left\{{}\begin{matrix}x^3-x+1=u\\u^3+1=2x\end{matrix}\right.\)
\(\Rightarrow\left(x^3-x+1\right)-\left(u^3+1\right)=u-2x\)
\(\Rightarrow x^3+x-u^3-u=0\)
\(\Rightarrow\left(x-u\right)\left(x^2+xu+u^2\right)+\left(x-u\right)=0\)
\(\Rightarrow\left(x-u\right)\left(x^2+xu+u^2+1\right)=0\)
\(\Rightarrow x-u=0\Rightarrow x=u\)
\(\Rightarrow x=\sqrt[3]{2x-1}\)
\(\Leftrightarrow x^3=2x-1\Leftrightarrow x^3-2x+1=0\)
\(\Leftrightarrow x^3-x^2+x^2-x-\left(x-1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Vậy phương trình (1) có các nghiệm là \(1;\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\)
Bài 16:
\(9x^3+2x-\sqrt[3]{3x+24}-8=0\left(1\right)\)
Đặt \(u=\sqrt[3]{3x+24}\Rightarrow u^3=3x+24\). Ta có hệ phương trình:
\(\left\{{}\begin{matrix}9x^3+2x-u-8=0\\u^3=3x+24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}27x^3+6x-3u-24=0\\u^3-3x-24=0\end{matrix}\right.\)
\(\Rightarrow\left(27x^3+6x-3u-24\right)-\left(u^3-3x-24\right)=0\)
\(\Rightarrow\left(27x^3-u^3\right)+3\left(3x-u\right)=0\)
\(\Rightarrow\left(3x-u\right)\left(9x^2+3xu+u^2\right)+3\left(3x-u\right)=0\)
\(\Rightarrow\left(3x-u\right)\left(9x^2+3xu+u^2+3\right)=0\)
\(\Rightarrow3x-u=0\Rightarrow3x=u\)
\(\Rightarrow3x=\sqrt[3]{3x+24}\)
\(\Leftrightarrow27x^3=3x+24\Leftrightarrow9x^3-x-8=0\)
\(\Leftrightarrow9x^3-9x^2+9x^2-9x+8x-8=0\)
\(\Leftrightarrow9x^2\left(x-1\right)+9x\left(x-1\right)+8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(9x^2+9x+8\right)=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy phương trình (1) có nghiệm duy nhất là 1.
15:
\(\Leftrightarrow x^3-x=\sqrt[3]{2x-1}-1\)
=>\(x\left(x-1\right)\left(x+1\right)=\dfrac{2x-1-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\)
=>x-1=0
=>x=1
