Question

ДАМ 35 БАЛЛОВ!
решите уравнение:
a.$6-\sqrt{6+x}=5$
b.$x+\sqrt{3x+7}=7$
c.$(16-x^{2})*\sqrt{3-x}=0$

Answer 1

a)

$6 - \sqrt{6 + x} = 5$

$\sqrt{6+x} = 1$

$6 + x = 1\\6 + x \geq 0$

$x = -5\\x \geq -6$

Ответ: -5.

б)

$x + \sqrt{3x+7} = 7$

$\sqrt{3x+7} = 7-x$

$3x+7 = (7-x)^2 \\3x + 7 \geq 0$

$3x + 7 = 49 - 14x + x^2\\x \geq -\frac{7}{3}$

$x^2 - 17x + 42 = 0$

$\{ {{x=3} \atop {x=14}} \right.$

Проверка:

$x_1 -$ $3 + \sqrt{9 + 7} = 3 + 4 = 7$

$x_2 -$ $14 + \sqrt{3*14 + 7} = 14 + \sqrt{49} = 14 + 7 = 21 \neq 7$

$x = 3$

Ответ: 3.

в)

$(16 - x^2) * \sqrt{3-x} = 0$

$\left \{ {{16-x^2 = 0} \atop {\sqrt{3-x} = 0}} \right.$

$\left \{ {{(4-x)(4+x) = 0} \atop {x = 3}} \atop {3-x\geq 0}\right.$

$\left[\begin{array}{cccc}x = 4\\x = -4\\x = 3\\x\leq3 \end{array}\right]$

$\left \{ {{x = -4} \atop {x=3}} \right.$

Ответ: -4; 3.