Rozwiąż równania:b)1+√3x =x+2c)2-√2x=x-4d)2x+3+x√3=4e)(x+2√2)√2=√3x-1f)√(x-√)=10-xg) 1 √3--------- =.... Question from @Kika96157

October 2018 1 13 Report

Rozwiąż równania:
b)1+√3x =x+2
c)2-√2x=x-4
d)2x+3+x√3=4
e)(x+2√2)√2=√3x-1
f)√(x-√)=10-x
g) 1 √3
--------- = -----
√3+x x

h) 2x-3 1
------- = ----
x√7 2

Jakubbartt

b) $1+\sqrt{3}x=x+2 \\ \sqrt{3}x-x=2-1 \\ x(\sqrt{3}-1)=1|:(\sqrt{3}-1) \\x=\frac{1}{(\sqrt{3}-1)}=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{(\sqrt{3}+1)}{3-1}=\frac{(\sqrt{3}+1)}{2}$

c) $2-\sqrt{2}x=x-4 \\-\sqrt{2}x-x=-6 \\-x(\sqrt{2}+1)=-6|:(-1) \\x(\sqrt{2}+1)=6|:(\sqrt{2}+1)]=\frac{6(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{6\sqrt{2}-6}{1}=6\sqrt{2}-6$

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

e) $(x+2\sqrt{2})\sqrt{2}=\sqrt{3}x-1 \\ \sqrt{2}x+4=\sqrt{3}x-1 \\ \sqrt{2}x-\sqrt{3}x= =-1-4 \\x(\sqrt{2}-\sqrt{3})=-5|:(\sqrt{2}-\sqrt{3})\\ x=\frac{-5}{(\sqrt{2}-\sqrt{3})} =\frac{-5(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}= (-5\sqrt{2}-5\sqrt{3}) :(-1)=</p>
<p>$

$\\= 5\sqrt{2}+5\sqrt{3}$

g) $\frac{1}{\sqrt{3}+x}=\frac{\sqrt{3}}{x} \\ x=3+\sqrt{3}x \\x-\sqrt{3}x=3 \\x(1-\sqrt{3})=3|:(1-\sqrt{3}) \\x=\frac{3}{(1-\sqrt{3})}=\frac{3(1+\sqrt{3})}{(1-\sqrt{3})(1-\sqrt{3})}=\frac{3+3\sqrt{3}}{1-3}=\frac{3+3\sqrt{3}}{-2}$

h) $\frac{2x-3}{x\sqrt{7}}=\frac{1}{2} \\x\sqrt{7} = 4x-6 \\x\sqrt{7}-4x=-6 \\x(\sqrt{7}-4)=-6 \\x=\frac{-6}{\sqrt{7}-4}=\frac{-6(\sqrt{7}+4)}{(\sqrt{7}-4)(\sqrt{7}+4)}=\frac{-6\sqrt{7}-24}{7-4}= \frac{-6\sqrt{7}-24}{3}=-2\sqrt{7}-8$