rozwiąż równania:a) 2x + 3 = 6∛5 + x *∛320b) 7 pierwiastków (5 stopnia) z -4 = -2 pierwiastki piąteg.... Question from @dancewiththedevil

dancewiththedevil @dancewiththedevil

October 2018 1 22 Report

rozwiąż równania:

a) 2x + 3 = 6∛5 + x *∛320

b) 7 pierwiastków (5 stopnia) z -4 = -2 pierwiastki piątego stopnia z 128 * x

c) 2x +√2 = 1 - √2 * x

Roma

$2x + 3 = 6\sqrt[3]{5}+ x \cdot \sqrt[3]{320} \\\ 2x - x \cdot \sqrt[3]{320} = 6\sqrt[3]{5} - 3 \\\ (2 - \sqrt[3]{5 \cdot 64}) \cdot x = 3 \cdot (2\sqrt[3]{5} - 1) \\\ (2 - 4\sqrt[3]{5}) \cdot x = 3 \cdot (2\sqrt[3]{5} - 1) \\\ - 2 \cdot (2\sqrt[3]{5} - 1)\cdot x = 3 \cdot (2\sqrt[3]{5} - 1) \ / : (2\sqrt[3]{5} - 1) \\\ -2x = 3 \ /:(-2) \\\ x = - 1,5$

$7 \sqrt[5]{-4} = -2\sqrt[5]{128} \cdot x \\\ 2\sqrt[5]{128} \cdot x = -7 \sqrt[5]{-4} \\\ 2\sqrt[5]{32 \cdot 4} \cdot x = -7 \sqrt[5]{-1 \cdot 4} \\\ 2\sqrt[5]{2^5 \cdot 4} \cdot x = -7 \sqrt[5]{-1} \cdot \sqrt[5]{4} \\\ 2\sqrt[5]{2^5} \cdot \sqrt[5]{4} \cdot x = -7 \sqrt[5]{-1} \cdot \sqrt[5]{4} \\\ 2 \cdot 2 \cdot \sqrt[5]{4} \cdot x = -7 \cdot (-1) \cdot \sqrt[5]{4} \\\ 4\cdot \sqrt[5]{4} \cdot x =7\cdot \sqrt[5]{4} \ / :\sqrt[5]{4} \\\ 4x = 7 \ /: 4 \\\ x = 1,75$

$2x +\sqrt{2} = 1 - \sqrt{2} \cdot x \\\ 2x + \sqrt{2} \cdot x = 1 - \sqrt{2} \\\ (2 + \sqrt{2}) \cdot x = 1 - \sqrt{2} \ /: (2 + \sqrt{2}) \\\ x = \frac{1 - \sqrt{2}}{2 + \sqrt{2}} \\\ x = \frac{(1 - \sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \\\ x = \frac{2 - \sqrt{2}-2\sqrt{2}+2}{4-2} \\\ x = \frac{4 - 3\sqrt{2}}{2}$