Usuń niewymierność z mianownika a) 3 - √2/ √6 b) 3+√2/√2 50. Usuń niewymierność z mianownika. a) 1/3.... Question from @WielkiKasztelan

October 2018 1 23 Report

Usuń niewymierność z mianownika a) 3 - √2/ √6 b) 3+√2/√2 50. Usuń niewymierność z mianownika. a) 1/3-√2 b) 4/√6 – 2 c) 2/ 3+√5 d) 2√3/√3+2 51. Usuń niewymierność z mianownika: a) 9/5√3 b) 2/√3 – 1 c) √2/3+√2 d) 7√2/2√2-3 e) √5 +1/ √5 -1 f) √6 / √2 + √3 g) √3 – 1/ 2√3 +√2 Zad 54. Rozwiąż równanie a) (2x-1)2 = (2x+3)2 b) (3x+2)2= (3x-4)(3x+4) c) (3x+1)2+(4x+1)2=(5x+1)2 d) (√2x-2)2 – 2(x-1)2= 4x Zad 55. Rozwiąż nierówność a) (x+3)2 – (x+2)2 ≤ 0 b) (4-x)2 + (2+x)2 > 2x2 c) 9x2 + (2+3x)(2-3x) ≥ 6x d) (3/2x – 4)(3/2x+4) < (3/2x – 1 )2 Zad 56. a) ( 6x -3 )2 ≤ (6x-1)2 b) (4x+3)2= (2x+5)(5-2x) c) (5x-2)2 – (3x -1 )2= (4x-1)(4x+1) d) 3(x+1)2-(√3x -√2)2= 6x

agulka1987

$a)\\ \frac{3 -\sqrt{2}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}-\sqrt{12}}{6}=\frac{3\sqrt{6}-2\sqrt{3}}{6}\\ b)\\ \frac{3+\sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}+2}{2}$

Zad.50

$a)\\ \frac{1}{3-\sqrt{2}} \cdot \frac{3+\sqrt{2}}{3+\sqrt{2}}=\frac{3+\sqrt{2}}{9-2}=\frac{3+\sqrt{2}}{7}\\ b)\\ \frac{4}{\sqrt{6}-2} \cdot \frac{\sqrt{6}+2}{\sqrt{6}+2}=\frac{4(\sqrt{6}+2)}{6-2}=\frac{4(\sqrt{6}+2)}{4}=\sqrt{6}+2\\ c)\\ \frac{2}{ 3+\sqrt{5 }} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{2(3-\sqrt{5})}{9-5}=\frac{3-\sqrt{5}}{2}\\ d)\\ \frac{2\sqrt{3}}{\sqrt{3}+2 }\cdot \frac{\sqrt{3}-2}{\sqrt{3}-2} = \frac{6-4\sqrt{3}}{3-4}=\frac{6-4\sqrt{3}}{-1}=-(6-4\sqrt{3})=4\sqrt{3}-6$

Zad.51

$a)\\ \frac{9}{5\sqrt{3}}\cdot \frac{ \sqrt{3}}{\sqrt{3}}=\frac{9\sqrt{3}}{15}=\frac{3\sqrt{3}}{5}\\ b)\\ \frac{2}{\sqrt{3}-1 } \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2(\sqrt{3}+1)}{3-1}=\sqrt{3}+1\\ c)\\ \frac{\sqrt{2}}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3\sqrt{2}-2}{9-2}=\frac{3\sqrt{2}-2}{7}\\ d)\\ \frac{7\sqrt{2}}{2\sqrt{2}-3} \cdot \frac{2\sqrt{2}+3}{2\sqrt{2}+3}=\frac{28+21\sqrt{2}}{8-9}=-(28+21\sqrt{2})\\$

$e)\\ \frac{\sqrt{5} +1}{\sqrt{5} -1} \cdot \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{(\sqrt{5}+1)^2}{5-1}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}\\ f)\\ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{12}-\sqrt{18}}{2-3}=\frac{2\sqrt{3}-3\sqrt{2}}{-1}=3\sqrt{2}-2\sqrt{3}\\ g)\\ \frac{\sqrt{3}-1}{2\sqrt{3} +\sqrt{2}} \cdot \frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}-\sqrt{2}}=\frac{6-\sqrt{6}-2\sqrt{3}+\sqrt{2}}{12-2}=\frac{6-\sqrt{6}-2\sqrt{3}+\sqrt{2}}{10}$

Zad.54

$a)\\ (2x-1)^2 = (2x+3)^2 \\ 4x^2-4x+1=4x^2+12x+9\\ -16x=8\\ x=-\frac{1}{2} b)\\ (3x+2)^2= (3x-4)(3x+4)\\ 9x^2+12x+4=9x^2-16\\ 12x=-20\\ x=-\frac{20}{12}=-\frac{5}{3}=-1\frac{2}{3}\\ c)\\ (3x+1)^2+(4x+1)^2=(5x+1)^2\\ 9x^2+6x+1+16x^2+8x+1=25x^2+10x+1\\ 4x=-1\\ x=-\frac{1}{4}\\ d)\\ (\sqrt{2}x-2)^2-2(x-1)^2= 4x\\ 2x^2-4\sqrt{2}x+4- 2x^2+4x-2=4x\\ -4\sqrt{2}x=-2\\ x=-\frac{2}{4\sqrt{2}}=-\frac{2\sqrt{2}}{8}=-\frac{\sqrt{2}}{4}$

Zad.55

$a)\\ (x+3)^2-(x+2)^2 \le 0\\ x^2+6x+9-x^2-4x-4 \le0\\ 2x \le -5\\ x \le -\frac{5}{2}\\ x \in (-\infty \ , \ -\frac{5}{2}>\\ b)\\ (4-x)^2 + (2+x)^2 > 2x^2\\ 16-8x+x^2+4+4x+x^2>2x^2\\ -4x>-20\\ x<5\\ x \in (-\infty, 5)\\ c)\\ 9x^2 + (2+3x)(2-3x) \ge 6x\\ 9x^2+4-9x^2 \ge 6x\\ 6x \le 4\\ x \le \frac{2}{3}\\ x \in (-\infty \ , \ \frac{2}{3}>\\ d)\\ (\frac{3}{2}x-4)(\frac{3}{2}x+4) < (\frac{3}{2}x -1 )^2\\ \frac{9}{4}x^2-16<\frac{9}{4}x^2-3x+1\\ -3x> -17\\ x<\frac{17}{3}\\ x \in (-\infty , \frac{17}{3})$

Zad.56

$a)\\ ( 6x -3 )^2 \le (6x-1)^2\\ 36x^2-36x+9 \le 36x^2-12x+1\\ -24x \le -8\\ x \ge \frac{1}{3}\\ x \in <\frac{1}{3}, +\infty)\\ b)\\ (4x+3)^2= (2x+5)(5-2x)\\ 16x^2+24x+9=25-4x^2\\ 20x^2-24x-16=0 \setminus :4\\ 5x^2-6x-4=0\\ \Delta=36+80=116\\ x_{1}=\frac{6-2\sqrt{29}}{10}=\frac{3-\sqrt{29}}{5}\\ x_{2}=\frac{6+2\sqrt{29}}{10}=\frac{3+\sqrt{29}}{5}\\ x=\{\frac{3-\sqrt{29}}{5} \ , \ \frac{3+\sqrt{29}}{5}\}\\ c)\\ (5x-2)^2 -(3x -1 )^2= (4x-1)(4x+1) \\ 25x^2-20x+4-9x^2+6x-1=16x^2-1\\ -14x=-4\\ x=\frac{2}{7}\\$