Usuń niewymierność z mianownika²/√₃Wykonaj działania(7-x)²=(x+½)²=(2x+4)²=(3x+8)²=(¼x+8)²=(⅔x-6)²=(0.... Question from @Adusia2518

September 2018 1 26 Report

Usuń niewymierność z mianownika
²/√₃ $\frac{2}{\sqrt{3}}= \\\frac{4}{3\sqrt{2}}= \\\frac{1}{6+\sqrt{2}}= \\\frac{2\sqrt{2}}{\sqrt{5}+4}= \\\frac{2}{4-3\sqrt{2}}= \\\frac{2\sqrt{3}-1}{2\sqrt{3}+2}= \\\frac{1}{\sqrt{3}-\sqrt{2}}=$

Wykonaj działania

(7-x)²=

(x+½)²=

(2x+4)²=

(3x+8)²=

(¼x+8)²=

(⅔x-6)²=

(0,5x+0,4)²=

(0,1x+2,5)²=

wykonaj działania, korzystając za wzorów skróconego mnożenia.

wzory:

1. (a+b)²=a²+2ab+b²

2. (a-b)²=a²-2ab+b²

3. a²-b²=(a-b)(a+b)

zadanie:

$(3+\sqrt{2})^{2}= \\(1-\sqrt{5})^{2}= \\(4-3\sqrt{2})^{2}= \\(2\sqrt{7}+3)^{2}= \\(2\sqrt{5}+\sqrt{3})^{2}= \\(\sqrt{6}-2\sqrt{3})^{2}= \\\sqrt{6}-\frac{\sqrt{2}}{2})^{2}= \\(\frac{\sqrt{6}}{2}+\frac{\sqrt{3}}{3}$ =

Oblicz wartość wyrażenia:

a. (x+1)(x-1)+(x+2)-(x-2)-(x+3)(x-3) dla x=√3

b. (1-2x)(1+2x)(1-3x)(1+3x)(1-4x)(4x+1) dla x=√5

Oblicz obwód trójkąta prostokątnego o przyprostokątnych a i b.

a. a=4-√2, b=4+√2

b. a=8+√2, b=4-2√2

cyfra

zadanie 1

$\frac{2}{\sqrt{3}}= \frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}= \frac{2\sqrt{3}}{3} \\ \frac{4}{3\sqrt{2}}= \frac{4}{3\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}= \frac{4\sqrt{2}}{3*2} = \frac{2\sqrt{2}}{3} \\ \frac{1}{6+\sqrt{2}}= \frac{1}{6+\sqrt{2}}*\frac{6-\sqrt{2}}{6-\sqrt{2}} = \frac{6 - \sqrt{2}}{6^2 - 2} = \frac{6 - \sqrt{2}}{34}\\ \frac{2\sqrt{2}}{\sqrt{5}+4}= \frac{2\sqrt{2}}{\sqrt{5}+4}*\frac{\sqrt{5}-4}{\sqrt{5}-4}= \frac{2\sqrt{2}(\sqrt{5}-4)}{5 - 4^2} = \frac{2\sqrt{2}(4 - \sqrt{5})}{11} \\ \frac{2}{4-3\sqrt{2}}= \frac{2}{4-3\sqrt{2}}*\frac{4+3\sqrt{2}}{4+3\sqrt{2}}= \frac{2(4+3\sqrt{2})}{4^2 - 3^2*2} = \frac{2(4+3\sqrt{2})}{- 2} = - (4+3\sqrt{2}) \\ \frac{2\sqrt{3}-1}{2\sqrt{3}+2}= \frac{2\sqrt{3}-1}{2\sqrt{3}+2}*\frac{2\sqrt{3}-2}{2\sqrt{3}-2}=\frac{(2\sqrt{3}-1)(2\sqrt{3}-2)}{2^2*3 - 2^2}= \frac{14 - 6\sqrt{3}}{8}= \frac{7 - 3\sqrt{3}}{4} \\ \frac{1}{\sqrt{3}-\sqrt{2}}= \frac{1}{\sqrt{3}-\sqrt{2}}*\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}= \frac{\sqrt{3}-\sqrt{2}}{3 - 2} = \sqrt{3}-\sqrt{2}$

zadanie 2

$(7-x)^2= = 49 - 14x + x^2\\ (x+½)^2=x^2 + x + \frac{1}{4}\\ (2x+4)^2=4x^2 + 16x + 16 = 4(x^2 + 4x +4)\\ (3x+8)^2=9x^2 + 48x + 64\\ (¼x+8)^2=\frac{1}{16}x^2 + 4x + 64\\ (⅔x-6)^2=\frac{4}{9}x^2 - 8x + 36\\ (0,5x+0,4)^2= 0,25x^2 + 0,4x + 0,16\\ (0,1x+2,5)^2=0,01x^2 + 0,5x + 6,25$

zadanie 3

$(3+\sqrt{2})^{2}=9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2} \\ (1-\sqrt{5})^{2}= 1 - 2\sqrt{5} + 5 = 2(3 - \sqrt{5}) \\ (4-3\sqrt{2})^{2}= 16 - 24\sqrt{2} + 18 = 34 - 24\sqrt{2} = 2(17 - 12\sqrt{2}) \\ (2\sqrt{7}+3)^{2}= 28 + 12\sqrt{7} + 9 = 37 + 12\sqrt{7} \\ (2\sqrt{5}+\sqrt{3})^{2}= 20 + 4\sqrt{15} + 3 = 23 + 4\sqrt{15} \\ (\sqrt{6}-2\sqrt{3})^{2}= 6 - 4\sqrt{18} + 12 = 18 - 12\sqrt{2} = 6(3 - 2\sqrt{2} \\ (\sqrt{6}-\frac{\sqrt{2}}{2})^{2}= 6 - \sqrt{12} + \frac{2}{4} = 6\frac{1}{2} - 2\sqrt{3} \\ (\frac{\sqrt{6}}{2}+\frac{\sqrt{3}}{3})^2 = \frac{6}{4} + \frac{\sqrt{18}}{3} + \frac{3}{9} = \frac{3}{2} + \frac{1}{3} + \frac{3\sqrt{2}}{3} = \frac{11}{6} + \sqrt{2}$

zadanie 4

$c = \sqrt{a^2 + b^2}\\ c_a = \sqrt{(4 - \sqrt{2})^2 + (4 + \sqrt{2})^2} = \sqrt{16 - 8\sqrt{2} + 2 + 16 + 8\sqrt{2} + 2} = \sqrt{36} = 6\\ D_a = 4 - \sqrt{2} + 4 + \sqrt{2} + 6 = 14\\ c_b = \sqrt{(8 + \sqrt{2})^2 + (4 - 2\sqrt{2})^2} = \sqrt{64 + 16\sqrt{2} + 2 + 16 - 16\sqrt{2} + 8} = \sqrt{90} = 3\sqrt{10} \\ D_b = 8 + \sqrt{2} + 4 - 2\sqrt{2} + 3\sqrt{10} = 12 - \sqrt{2} + 3\sqrt{3}$