Question

[Thế Nam]

\(^{x^2-x+2\sqrt{x^3+1}=2\sqrt{x+1}}\)

\(5\sqrt{x^3+8}=2\left(x^2-x+6\right)\)

Answer 1

[Lê Song Phương]

a) đkxđ \(x\ge-1\)

pt đã cho tương đương với

\(x^2-x=2\left(\sqrt{x+1}-\sqrt{x^3+1}\right)\)

\(\Leftrightarrow x^2-x=2.\dfrac{x+1-\left(x^3+1\right)}{\sqrt{x+1}+\sqrt{x^3+1}}\)

\(\Leftrightarrow x\left(x-1\right)=2.\dfrac{x\left(1-x\right)}{\sqrt{x+1}+\sqrt{x^3+1}}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(nhận\right)\\1+\dfrac{1}{\sqrt{x+1}+\sqrt{x^3+1}}=0\left(vôlí\right)\end{matrix}\right.\)

Vậy pt đã cho có tâp nghiệm \(S=\left\{0;-1\right\}\)

Answer 2

[Nguyễn Đức Trí]

\(x^2-x+2\sqrt[]{x^3+1}=2\sqrt[]{x+1}\) 

\(\Leftrightarrow2\sqrt[]{x^3+1}-2\sqrt[]{x+1}-\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}=0\)

\(\Leftrightarrow2\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)-\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=2\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)-\dfrac{1}{4}\left(1\right)\)

mà \(\left(x+\dfrac{1}{2}\right)^2\ge0,\forall x\inℝ\)

\(\left(1\right)\Leftrightarrow2\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)-\dfrac{1}{4}\ge0\)

\(\Leftrightarrow\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)\ge\dfrac{1}{8}\left(2\right)\)

Điều kiện xác định :

\(\left\{{}\begin{matrix}x+1\ge0\\\sqrt[]{x^2-x+1}-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\sqrt[]{x^2-x+1}\ge1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x+1\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\left(x-1\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le0\cup x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\-1\le x\le0\end{matrix}\right.\)

BPT \(\left(2\right)\Leftrightarrow\left(x+1\right)\left(x^2-x+1-2\sqrt[]{x^2-x+1}-1\right)\ge\dfrac{1}{64}\)

\(\Leftrightarrow\left(x^2-x-2\sqrt[]{x^2-x+1}\right)\ge\dfrac{1}{64}\left(vì.x+1\ge0\right)\)

Đặt \(t=\sqrt[]{x^2-x+1}>0\)

\(BPT\Leftrightarrow t^2-2t-1-\dfrac{1}{64}\ge0\)

\(\Leftrightarrow t^2-2t-\dfrac{63}{64}\ge0\)

\(\Leftrightarrow t^2-2t+1-1-\dfrac{63}{64}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2-\dfrac{127}{64}\ge0\)

\(\Leftrightarrow\left(t-1-\dfrac{\sqrt[]{127}}{8}\right)\left(t-1+\dfrac{\sqrt[]{127}}{8}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}t\ge1+\dfrac{\sqrt[]{127}}{8}\\t\le1-\dfrac{\sqrt[]{127}}{8}\end{matrix}\right.\)

\(\Leftrightarrow t\ge1+\dfrac{\sqrt[]{127}}{8}\)  \(\left(t>0;1-\dfrac{\sqrt[]{127}}{8}< 0\right)\)

\(\Leftrightarrow\sqrt[]{x^2-x+1}\ge1+\dfrac{\sqrt[]{127}}{8}\)

\(\Leftrightarrow x^2-x+1\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2\)

mà \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\)

      \(\dfrac{3}{4}< \left(1+\dfrac{\sqrt[]{127}}{8}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\le-\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}\\x-\dfrac{1}{2}\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\\x\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\) (so với đkxđ \(\left[{}\begin{matrix}x\ge1\\-1\le x\le0\end{matrix}\right.\))

\(\Leftrightarrow x=\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\)