Giải phương trình a, $\sqrt[]{2-x}$ +$\sqrt[]{2+x}$ + 3$\sqrt[]{4-x^{2}}$ = 2 b, 2($x^{2}$+2)=5($x^{3}$ +1)

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1. Đáp án: $a)x=±2$

                `b)x=\frac{5±\sqrt{37}}{2}`

   Giải thích các bước giải:

   $a)ĐKXĐ:-2≤x≤2$

   $\sqrt{2-x}+\sqrt{2+x}+3\sqrt{4-x^2}=2$

   $⇔\sqrt{2-x}+\sqrt{2+x}=2-3\sqrt{4-x^2}$

   $⇔(2-x)+(2+x)+2\sqrt{2-x}.\sqrt{2+x}=4+9(4-x^2)-12\sqrt{4-x^2}$

   $(ĐK:\left[ \begin{array}{l}-2≤x≤\frac{-\sqrt{7}}{2}\\\frac{\sqrt{7}}{2}≤x≤2\end{array} \right.)$

   $⇔2\sqrt{4-x^2}=9(4-x^2)-12\sqrt{4-x^2}$

   $⇔9(4-x^2)-14\sqrt{4-x^2}=0$

   $⇔\sqrt{4-x^2}(9\sqrt{4-x^2}-14)=0$

   $⇔\left[ \begin{array}{l}\sqrt{4-x^2}=0(1)\\9\sqrt{4-x^2}-14=0(2)\end{array} \right.$

   $\text{Ta có:}$

   $(1)⇔4-x^2=0⇔x^2=4⇔x=±2$ $\text{(thỏa mãn ĐKXĐ)}$

   `(2)⇔\sqrt{4-x^2}=\frac{14}{9}`

   `⇔4-x^2=\frac{196}{81}`

   `⇔x^2=\frac{128}{81}`

   `⇔x=±\frac{8\sqrt{2}}{9}` $\text{(không thỏa mãn ĐKXĐ)}$

   $\text{Vậy phương trình có nghiệm x=±2}$

   $b)ĐKXĐ:x≥-1$

   $\text{Đặt}$$\sqrt{x+1}=a;\sqrt{x^2-x+1}=b(a≥0;b>0)$

   $⇒x+1=a^2;x^2-x+1=b^2$

   $\text{Ta có:}$

   $2[(x^2-x+1)+(x+1)]=5.\sqrt{x+1}.\sqrt{x^2-x+1}$

   $⇔2(a^2+b^2)=5ab$

   $⇔2a^2-5ab+2b^2=0$

   $⇔(2a-b)(a-2b)=0$

   $⇔\left[ \begin{array}{l}2a-b=0(3)\\a-2b=0(4)\end{array} \right.$ 

   $\text{Ta có:}$

   $(3)⇔2a=b$

   $⇔2\sqrt{x+1}=\sqrt{x^2-x+1}$

   $⇔\sqrt{4x+4}=\sqrt{x^2-x+1}$

   $⇔4x+4=x^2-x+1$

   $⇔x^2-5x-3=0$

   `⇔(x^2-2.x.\frac{5}{2}+\frac{25}{4})-\frac{37}{4}=0`

   `⇔(x-\frac{5}{2})^2=\frac{37}{4}`

   `⇔x-\frac{5}{2}=\frac{±\sqrt{37}}{2}`

   `⇔x=\frac{5±\sqrt{37}}{2}` $\text{(thỏa mãn ĐKXĐ)}$

   $(4)⇔a=2b$

   $⇔\sqrt{x+1}=2\sqrt{x^2-x+1}$

   $⇔\sqrt{x+1}=\sqrt{4x^2-4x+4}$

   $⇔x+1=4x^2-4x+4$

   $⇔4x^2-5x+3=0$

   `⇔(4x^2-2.2x.\frac{5}{4}+\frac{25}{16})+\frac{23}{16}=0`

   `⇔(2x-\frac{5}{4})^2=\frac{-23}{16}` $\text{(vô lý)}$

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