Question

Решите уравнения с корнями:

Answer 1

$1)\sqrt{13+\sqrt{11-\sqrt{2x+1} } } =4\\13+\sqrt{11-\sqrt{2x+1}} =16\\\sqrt{11-\sqrt{2x+1}}=3\\11-\sqrt{2x+1}=9,\\-\sqrt{2x+1}=-2,\\\sqrt{2x+1}=2\\2x+1=4,\\2x=3\\x=1,5.$

$2)\sqrt{3x+4}=\sqrt{8-x}\\3x+4=8-x\\ 3x+x=8-4\\4x=4\\x=1$

$3) \sqrt{1-x}=\sqrt{2x-6}\\OD3: 2x - 6 \geq 0 \Rightarrow x \geq 3\\ 1-x=2x-6\\2x+x=1+6\\3x=7\\x=\frac{7}{3} \notin OD3$

ОТВЕТ: 1)1,5; 2)1; 3) нет корней.

Answer 2

1) $\sqrt{13+\sqrt{11-\sqrt{2x+1}}}=4=>13+\sqrt{11-\sqrt{2x+1}}=16=>\sqrt{11-\sqrt{2x+1}}=3=>11-\sqrt{2x+1}=9=>\sqrt{2x+1}=2=>2x+1=4=>2x=3=>x=1\dfrac{1}{2}$

2) $\sqrt{3x+4}=\sqrt{8-x}=>\left \{ {{3x+4\geq 0} \atop {8-x\geq0 }} \right. =>\left \{ {{x\geq -\frac{4}{3}} \atop {8\geq x }} \right. \\ 3x+4=8-x=>4x=4=>x=1\\ -\frac{4}{3}\leq 1\leq 8=>x=1$

3) $\sqrt{1-x}=\sqrt{2x-6}=>\left \{ {{1-x\geq 0} \atop {2x-6\geq 0 }} \right. =>\left \{ {{1\geq x} \atop {x\geq 3 }} \right. =>x\in \o$

=> Нет корней