Условия задач

1. $x|2x+1|-3x-4=0$
2. $x^2-5x\cdot\frac{|x-2|}{x-2}-14=0$
3. $(8x^2-3x+1)^2=32x^2-12x+1$
4. $\frac{(x^4-5x^2+4)(x^4+7x^2-18)}{(x-4)^2(3x-5)}\geq 0$
5. $\frac{x^2-7|x|+10}{x^2-6x+9}<0$
6. $\sqrt{x+7}\cdot\sqrt{3x-2}=3\sqrt{x-1}\cdot\sqrt{x+2}$
7. $\sqrt{1+x\sqrt{x^2+42}}=x+1$
8. $\sqrt{2x+3}+\sqrt{3x+2}=\sqrt{2x+5}+\sqrt{3x}$
9. $\sqrt{4-\sqrt{1-x}}-\sqrt{2-x}>0$
10. $\left\{\begin{array}{l l} (x+y+1)^2+(x+y)^2=25,\\ x^2-y^2=3\end{array}\right.$
11. $\left\{\begin{array}{l l} \sqrt{\frac{y}{x}}-2\sqrt{\frac{x}{y}}=2,\\ \sqrt{5x+y}+\sqrt{5x-y}=4 \end{array}\right.$

Ответ

1. 2
2. -7; 7
3. 0; 3/8; $(3\pm\sqrt{73})/16$
4. $[-2;-\sqrt{2}]\cup [-1;1]\cup [\sqrt{2};5/3)\cup [2;4)\cup (4;+\infty)$
5. $(-5; -2)\cup (2;3)\cup (3;5)$
6. 2
7. 9,5; 0
8. 3
9. $(\frac{\sqrt{13}-5}{2}; 1]$
10. (-19/8; -13/8), (2;1)
11. (1;4)