Question

СРОЧНО!! решите уравнение:​

Answer 1

Обяснение:

1.

$\sqrt{x+1} =3$

ОДЗ: х+1≥0       х≥-1.

$(\sqrt{x+1} )^2=3^2\\\\x+1=9\\\\x=8.$

Ответ: х=8.

2.

$\sqrt{2x+3} =x$

ОДЗ:

$\displaystyle\\\left \{ {{2x+3\geq 0} \atop {x\geq 0}} \right. \ \ \ \ \ \ \left \{ {{2x\geq -3} \atop {x\geq 0}} \right.\ \ \ \ \ \ \left \{ {{x\geq -1,5} \atop {x\geq 0}} \right.\ \ \ \ \Rightarrow \ \ \ \ x\geq 0.$

$(\sqrt{2x+3})^2 =x^2\\\\2x+3=x^2\\\\x^2-2x-3=0\\\\x^2-3x+x-3=0\\\\x*(x-3)+(x-3)=0\\\\(x-3)*(x+1)=0\\\\x-3=0\\\\x_1=3.\\\\x+1=0\\\\x=-1\notin.$

Ответ: x=3.

3.

$\sqrt{-4x^2-16}=2$

ОДЗ:

$-4x^2-16\geq 0\ |:(-4)\\\\x^2+4\leq 0\ \ \ \ \Rightarrow\ \ \ \ x\in \varnothing.$

Ответ: уравнение не имеет решения.

4.

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

ОДЗ:

$\displaystyle\\\left \{ {{8-4x\geq 0} \atop {x+1\geq 0}} \right. \ \ \ \ \left \{ {{4x\leq 8\ |:4} \atop {x\geq -1}} \right. \ \ \ \ \left \{ {{x\leq 2} \atop {x\geq -1}} \right. \ \ \ \ \Rightarrow\ \ \ \ x\in[-1;2].$

$(x+1)^2=(\sqrt{8-4x} )^2\\\\x^2+2x+1=8-4x\\\\x^2+6x-7=0\\\\x^2+7x-x-7=0\\\\x*(x+7)-(x+7)=0\\\\(x+7)*(x-1)=0\\\\x+7=0\\\\x_1=-7\notin.\\\\x-1=0\\\\x_2=1.$

Ответ: х=1.

5.

$\sqrt{2x}+\sqrt{x-3} =-1\\\\\sqrt{2x}\geq 0\ \ \ \ \ \ \sqrt{x-3}\geq 0\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ \sqrt{2x}+\sqrt{x-3}\geq 0\\\\ -1 < 0\ \ \ \ \ \ \Rightarrow$

Ответ: уравнение решения не имеет.