$x+\sqrt[3]{x+1}+\sqrt[3]{x-1}+3\sqrt[3]{x^3-x^2}=\sqrt[3]{x}+3\sqrt[3]{x^3-2x^2+x}\space \space (1)$ $(1)\Leftrightarrow \left ( \sqrt[3]{x+1}+\sqrt[3]{x-1}-\sqrt[3]{x} \right )\left[ {{{\left ( \sqrt[3]{x+1}-\frac{\sqrt[3]{x-1}-\sqrt[3]{x}}{2} \right )}^{2}}+\frac{3{{( \sqrt[3]{x-1}-\sqrt[3]{x} )}^{2}}}{4}+1} \right]=0$$\Leftrightarrow \sqrt[3]{x+1}+\sqrt[3]{x-1}-\sqrt[3]{x}=0$Bài toán được giải quyết

$x+\sqrt[3]{x+1}+\sqrt[3]{x-1}+3\sqrt[3]{x^3-x^2}=\sqrt[3]{x}+3\sqrt[3]{x^3-2x^2+x}\space \space (1)$ $(1)\Leftrightarrow \left ( \sqrt[3]{x+1}+\sqrt[3]{x-1}-\sqrt[3]{x} \right )\left[ {{{\left ( \sqrt[3]{x+1}-\frac{\sqrt[3]{x-1}-\sqrt[3]{x}}{2} \right )}^{2}}+\frac{3{{( \sqrt[3]{x-1}-\sqrt[3]{x} )}^{2}}}{4}+1} \right]=0$$\Leftrightarrow \sqrt[3]{x+1}+\sqrt[3]{x-1}-\sqrt[3]{x}=0$$\Leftrightarrow \sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{x}\space \space (2)$$x=0$ là nghiệm của (2)$x\neq 0$ chia hai vế của (2) cho $\sqrt[3]{x}$ ta được :$\sqrt[3]{\frac{x+1}{x}}+\sqrt[3]{\frac{x-1}{x}}=1$Đặt $\left\{ \begin{array}{l} a=\sqrt[3]{\frac{x+1}{x}}\\ b=\sqrt[3]{\frac{x-1}{x}} \end{array} \right.$Ta thu được : $\left\{ \begin{array}{l} a+b=1\\ a^3+b^3=2 \end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} b=1-a\\ 3a^2-3a-1=0 \end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} a=\frac{3-\sqrt{21}}{6} \\ b=\frac{3+\sqrt{21}}{6} \end{array} \right.$ $\vee \left\{ \begin{array}{l} a=\frac{3+\sqrt{21}}{6} \\ b=\frac{3-\sqrt{21}}{6} \end{array} \right.$$\Leftrightarrow x=\frac{3\sqrt{21}}{14} \vee x=-\frac{3\sqrt{21}}{14}$Thử lại thấy : $x=0; x=\frac{3\sqrt{21}}{14}; x=-\frac{3\sqrt{21}}{14}$ thoả (1)Vậy phương trình có ba nghiệm : $x=0; x=\frac{3\sqrt{21}}{14}; x=-\frac{3\sqrt{21}}{14}$Bài toán được giải quyết.

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\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)=0$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$

\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)=0$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$

\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)=0$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$

\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)=0$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$

\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$

\begin{cases}2x^2+5x+4=2y+\sqrt{y+2x+3} \space \space (1)\\ y-3x-6-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (2)\end{cases}Điều kiện : $\begin{cases}x\ge2 \\ y+2x+3\ge0 \end{cases}$ (*)$(1)\Leftrightarrow (2x+5+2\sqrt{y+2x+3})(x+2-\sqrt{y+2x+3})=0$$\Leftrightarrow x+2=\sqrt{y+2x+3}$ vì $2x+5+2\sqrt{y+2x+3}>0 \space \forall$ x, y thoả (*)$\Leftrightarrow y=x^2+2x+1$Thay vào (2) ta được : ${{x}^{2}}-x-5-5\sqrt{2x-1}+2\sqrt{x-2}-2\sqrt[4]{2x-1}=0 \space \space (3)$$(3)\Leftrightarrow \left[ \frac{\left( 2x-1 \right)\sqrt[4]{2x-1}}{2}+\sqrt[4]{{{\left( 2x-1 \right)}^{3}}}+\frac{7\sqrt[4]{2x-1}}{2}+2+\frac{\left( 2x+6+\sqrt{2x-1} \right)\sqrt{x-2}}{4} \right]\left( \sqrt{x-2}-\sqrt[4]{2x-1} \right)=0$$\Leftrightarrow \sqrt{x-2}=\sqrt[4]{2x-1}$ vì [...] > 0 $\forall x\ge2$$\Leftrightarrow x=5 \Rightarrow y=36$Vậy hệ có nghiệm duy nhất $(5;36)$