Question

Пожалуйста с1 и если сможете с2

Answer 1

С1:
$\sqrt{17-12 \sqrt{2}}+\sqrt{9-4 \sqrt{2}}=\sqrt{(9+8)-2*3*2 \sqrt{2}}+\sqrt{(8+1)-2*1*2 \sqrt{2}}=\sqrt{3^{2}-2*3*2 \sqrt{2}+(2 \sqrt{2})^{2}}+\sqrt{(2 \sqrt{2})^{2}-2*1*2 \sqrt{2}+1}=$$=\sqrt{(3-2 \sqrt{2})^{2}}+\sqrt{(2 \sqrt{2}-1)^{2}}=|3-2 \sqrt{2}|+|2 \sqrt{2}-1|=3-2 \sqrt{2}+2 \sqrt{2}-1=2$

C2:
ОДЗ:
$x+1\ \textgreater \ 0$
$\sqrt{x+1}-4 \neq 0$
$2+ \sqrt{x+1} \neq 0$

$\left \{ {{x\ \textgreater \ -1} \atop {x \neq 15}} \right.$

Решение:

Замена: $\sqrt{x+1}=t>0$
$x+1=t^{2}$
$x-15=(x+1)-16=t^{2}-16=(t-4)(t+4)$
$x-3=(x+1)-4=t^{2}-4=(t-2)(t+2)$

$\frac{(t-4)(t+4)}{t-4}+ \frac{(t-2)(t+2)}{t+2}=6$
$t+4+t-2=6$
$2t=4$
$t=2$

Вернемся к замене:
$\sqrt{x+1}=2$
$x+1=4$
$x=3$ - удовлетворяет ОДЗ

Ответ: 3

Answer 2

C1
$...= \sqrt{(3-2 \sqrt{2})^2 } + \sqrt{(2 \sqrt{2}-1)^2 }=3-2 \sqrt{2}+2 \sqrt{2} -1=2$
C2
√(x+1)≠4⇒x+1≠16⇒x≠15
...=(x-15)(√(x+1) +4)/(√(x+1)-4)(√(x+1) +4) +(x-3)(2-√(x+1))/(2+√(x+1))(2-√(x+1))=
=(x-15)(√(x+1) +4)/(x-15)+(x-3)(2-√(x+1))/(3-x)=6
(√(x+1) +4)-(2-√(x+1))=6
√(x+1)+4-2+√(x+1)=6
2√(x+1)=4
√(x+1)=2
x+1=4
x=3