Rozwiąż równania:a) 1 + √ 3x = x + 2b) 2 - √ 2x = x - 4c) (x+√ 3)(2+√ 3) = 1 + 3√ 3d) (x + 2√ 2)√ 2 .... Question from @krynio99

krynio99 @krynio99

November 2018 1 26 Report

Rozwiąż równania:

a) 1 + √ 3x = x + 2

b) 2 - √ 2x = x - 4

c) (x+√ 3)(2+√ 3) = 1 + 3√ 3

d) (x + 2√ 2)√ 2 = √ 3 - 1

e) (x - √ 5)(x + √ 5) = 1

f) (x + √ 2)^2 + x = x^2 + 3

g) 2^17 * x - 16^4 * 3 = 5 * (4^8 * x - 3 * 2^17)

h) 27^3 * 2x - 3^9 = 3^10 * x + 2 * 3^9

i) 3^13 * x - 4 * 9^6 = 27^4 * (2x - 5)

j) (4^5 * x + 32^2) * 2^5 = 2^16 * x

DAM NAJ !!!!

Hazyr

$a)\ 1+\sqrt{3}x=x+2=>\sqrt{3}x-x=2-1\\ x(\sqrt{3}-1)=1\ \ \ | :(\sqrt{3}-1)\\ x=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}\\ b)\ 2-\sqrt{2}x=x-4=>-\sqrt{2}x-x=-6\\ -x(\sqrt{2}+1)=-6\ \ \ \ | : (\sqrt{2}+1)\\ -x=\frac{-6}{\sqrt{2}+1}\ \ \ \ x=-6+6\sqrt{2}\\ c)\ (x+\sqrt{3})(2+\sqrt{3})=1+3\sqrt{3}=>2x+\sqrt{3}x+2\sqrt{3}+2=3\sqrt{3}\\ x(2+\sqrt{3})=\sqrt{3}-2\ \ \ \ | :(2+\sqrt{3})\\ x=\frac{\sqrt{3}-2}{2+\sqrt{3}}\ \ \ \ x=-7+4\sqrt{3}\\ d)\ (x+2\sqrt{2})*\sqrt{2}=\sqrt{3}-1=>\sqrt{2}x+4=\sqrt{3}-1\\$ $\sqrt{2}x=\sqrt{3}-5\ \ \ | :\sqrt{2}=>x=\frac{\sqrt{6}-5\sqrt{2}}{2}\\ e)\ (x-\sqrt{5})(x+\sqrt{5})=1=>x^2-5=1\ x^2=6\ \ \ | \sqrt{}\\ x=\sqrt{6}\ lub\ x=-\sqrt{6}\\ f)\ (x+\sqrt{2})^2+x=x^2+3=>x^2+2\sqrt{2}x+2+x=x^2+3\\ 2\sqrt{2}x+x=1=>x(2\sqrt{2}+1)=1.........x=\frac{-1+2\sqrt{2}}{7}\\ g)\ 2^1^7x-16^4x*3=5*(4^8x-3*2^1^7)-- nie\ potrafie\\ j)\ (4^5x+32^2)*2^5=2^1^6x=>2^5(2^1^0x+2^1^0)=2^1^6x\\ 2^1^5x+2^1^5=2^1^6x=>2^1^5x-2^1^6x=-2^1^5\\ x(2^1^5-2^1^6)=-2^1^5\ \ \ | : (2^1^5-2^1^6)\\ x=1$