Условия задач
- \(x|2x+1|-3x-4=0\)
- \(x^2-5x\cdot\frac{|x-2|}{x-2}-14=0\)
- \((8x^2-3x+1)^2=32x^2-12x+1\)
- \(\frac{(x^4-5x^2+4)(x^4+7x^2-18)}{(x-4)^2(3x-5)}\geq 0\)
- \(\frac{x^2-7|x|+10}{x^2-6x+9}<0\)
- \(\sqrt{x+7}\cdot\sqrt{3x-2}=3\sqrt{x-1}\cdot\sqrt{x+2}\)
- \(\sqrt{1+x\sqrt{x^2+42}}=x+1\)
- \(\sqrt{2x+3}+\sqrt{3x+2}=\sqrt{2x+5}+\sqrt{3x}\)
- \(\sqrt{4-\sqrt{1-x}}-\sqrt{2-x}>0\)
- \(\left\{\begin{array}{l l} (x+y+1)^2+(x+y)^2=25,\\ x^2-y^2=3\end{array}\right.\)
- \(\left\{\begin{array}{l l} \sqrt{\frac{y}{x}}-2\sqrt{\frac{x}{y}}=2,\\ \sqrt{5x+y}+\sqrt{5x-y}=4 \end{array}\right.\)
Ответ
- 2
- -7; 7
- 0; 3/8; \((3\pm\sqrt{73})/16\)
- \([-2;-\sqrt{2}]\cup [-1;1]\cup [\sqrt{2};5/3)\cup [2;4)\cup (4;+\infty)\)
- \((-5; -2)\cup (2;3)\cup (3;5)\)
- 2
- 9,5; 0
- 3
- \((\frac{\sqrt{13}-5}{2}; 1]\)
- (-19/8; -13/8), (2;1)
- (1;4)