$4\sqrt{x+2} + \sqrt{22-3x}= x^{2} + 8$

Bài toán:
Giải phương trình $$4\sqrt{x+2} + \sqrt{22-3x}= x^{2} + 8$$


Lời giải:

ĐK: $-2\leq x \leq \dfrac{22}{3}$

Cách 1:
PT tương đương:

$$4\left(\sqrt{x+2}-\dfrac{x+4}{3}\right)+\left(\sqrt{22-3x}-\dfrac{14-x}{3}\right)=x^2-x-2$$

$$\Leftrightarrow 4\left(\dfrac{9(x+2)-x^2-8x-16}{9\left(\sqrt{x+2}+\dfrac{x+4}{2}\right)}\right)+\left(\dfrac{9(22-3x)-x^2+28x-196}{9\left(\sqrt{22-3x}+\dfrac{14-x}{3}\right)}\right)=x^2-x-2$$

$$\Leftrightarrow \left( {{x^2} - x - 2} \right)\left( {1 + \dfrac{4}{{9\left( {\sqrt {x + 2}  + \dfrac{{x + 4}}{3}} \right)}} + \dfrac{1}{{9\left( {\sqrt {22 - 3x}  + \dfrac{{14 - x}}{3}} \right)}}} \right) = 0$$

$$\Leftrightarrow \left[ \begin{array}{l} x=-1\\ x = 2 \end{array} \right.$$

Cách 2:

$$\Leftrightarrow 36\sqrt {x + 2}  + 9\sqrt {22 - 3x}  = 9{x^2} + 72$$

$$\Leftrightarrow 9{x^2} - 9x - 18 + 12\left( {x + 4 - 3\sqrt {x + 2} } \right) + \left( {42 - 3x - 9\sqrt {22 - 3x} } \right) = 0$$

$$\Leftrightarrow 9\left( {{x^2} - x - 2} \right) + 12\frac{{{x^2} - x - 2}}{{x + 4 + 3\sqrt {x + 2} }} + \frac{{9{x^2} - 9x - 18}}{{42 - 3x + 9\sqrt {22 - 3x} }} = 0$$

$$\Leftrightarrow (x^2-x-2)\left [ 9+\frac{12}{x+4+3\sqrt{x+2}}+\frac{9}{42-3x+9\sqrt{22-3x}} \right ]=0$$
$$\Leftrightarrow \begin{bmatrix}x=-1  &  & \\ x=2  &  &  \end{bmatrix}$$

Cách 3:

Đặt $\sqrt {x + 2}=a$  với  $0 \leq a \leq \sqrt{\frac{28}{3}}$

$$PT \Leftrightarrow \sqrt {28 - 3{a^2}}  = {a^4} - 4{a^2} - 4a + 12$$

$$\Leftrightarrow \left( {6 - a} \right) - \sqrt {28 - 3{a^2}}  + \left( {{a^4} - 4{a^2} - 3a + 6} \right) = 0$$

$$\Leftrightarrow \frac{{4\left( {{a^2} - 3a + 2} \right)}}{{\left( {6 - a} \right) + \sqrt {28 - 3{a^2}} }} + \left( {{a^2} - 3a + 2} \right)\left( {{a^2} + 3a + 3} \right) = 0$$

$$\Leftrightarrow (a^2-3a+2)\left ( \frac{4}{(6-a)+\sqrt{28-3a^2}}+a^2+3a+3 \right )=0$$

$$\Leftrightarrow \begin{bmatrix}a=2  &  & \\ a=1  &  &  \end{bmatrix}$$

$$\Leftrightarrow \begin{bmatrix}x=-1  &  & \\ x=2  &  &  \end{bmatrix}$$