Question

решить, желательно с подробным решением ​

Answer 1

пишу неравенство

надеюсь

Answer 2

#1.

a)

$\sqrt{x^4 + 19} = 10\\x^4 + 19 = 100\\x^4 = 100 - 19\\x^4 = 81\\x = \sqrt[4]{81} \\x = 3$

б)

$\sqrt{1-2x} - 3 = \sqrt{16+x} \\1 - 2x -6\sqrt{1-2x} + 9 = 16 + x\\3x = -6(1+\sqrt{1-2x} )\\x = -2(1+\sqrt{1-2x} )\\x + 2 =-2\sqrt{1-2x} \\x^2 + 4x + 4 = 4(1-2x)\\x^2 + 4x + 4 = 4 - 8x \\x^2 + 12x = 0\\x(x+12) = 0\\x_1=0\\x_2 = -12$

#2.

$\left[\begin{array}{ccc}\sqrt{6+x} -3\sqrt{3y+4}=-10 \\4\sqrt{3y+4}-5\sqrt{6+x} = 6\end{array}\right\\$

$\sqrt{6+x} = 3\sqrt{3y + 4} -10$

$4\sqrt{3y+4} - 5(3\sqrt{3y+4} -10) = 6\\4\sqrt{3y+4} - 15\sqrt{3y+4} + 50 - 6 =0\\11\sqrt{3y+4} = 44\\\sqrt{3y+4} = 4\\3y+4 = 16\\3y = 12\\y = 4$

$\sqrt{6+x} = 3\sqrt{3*4+4} - 10\\\sqrt{6+x} = 3 * 4 - 10\\\sqrt{6+x} = 2\\6+x = 4\\x = -2$

ответ: $x = -2; y = 4.$

#3.

a)

$\sqrt{5-x} < 4\\\left \{ {{5-x\geq0 } \atop {5-x-11}} \right.$

x ∈ (-11; 5]

б)

$\sqrt{2x-5} < \sqrt{5x+4}$

$\left[\begin{array}{ccc}2x -5 < 5x + 4\\2x - 5 \geq 0\\5x + 4 \geq 0\end{array}\right$

$\left[\begin{array}{ccc}3x -9\\2x \geq 5\\5x \geq -4\end{array}\right$

$\left[\begin{array}{ccc}x -3\\x \geq \frac{5}{2} \\x \geq \frac{-4}{5} \end{array}\right$

x ∈ [2,5; +∞)