giải pt \(x^3+\sqrt{\left(1-x^2\right)^3}=x\sqrt{2\left(1-x^2\right)}\)
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18 tháng 6 2021
Đk:\(x\ge-1\)
Đặt \(\left(a,b,c\right)=\left(x;\sqrt{x+1};\sqrt{2}\right)\)
Pt tt: \(a^3+b^3+c^3=\left(a+b+c\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(\Leftrightarrow0=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(\Leftrightarrow3\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\a+c=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x+1}=0\\\sqrt{x+1}+\sqrt{2}=0\left(vn\right)\\x+\sqrt{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=-x\\x=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)\(\Rightarrow\)\(\sqrt{x+1}=-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le0\\x+1=x^2\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{5}}{2}\) (tm)
Vậy...
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ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1-x^2}=a\ge0\) ta được:
\(\left\{{}\begin{matrix}x^2+a^2=1\\x^3+a^3=\sqrt{2}ax\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(x^2+a^2-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(1-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
Đặt \(x+a=t\Rightarrow x^2+a^2+2ax=t^2\Rightarrow ax=\frac{t^2-1}{2}\)
\(\Rightarrow t\left(1-\frac{t^2-1}{2}\right)=\sqrt{2}\left(\frac{t^2-1}{2}\right)\)
\(\Leftrightarrow t^3+\sqrt{2}t^2-3t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+2\sqrt{2}t+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\\t=1-\sqrt{2}\\t=-1-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\sqrt{1-x^2}=\sqrt{2}\\x+\sqrt{1-x^2}=-1-\sqrt{2}\left(l\right)\\x+\sqrt{1-x^2}=1-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x^2}=\sqrt{2}-x\\\sqrt{1-x^2}=1-\sqrt{2}-x\left(x\le1-\sqrt{2}\right)\\\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x^2=\left(\sqrt{2}-x\right)^2\\1-x^2=\left(1-\sqrt{2}-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow...\)