a)
√2x - 5 = 0
\sqrt2x=5
x=\frac{5}{\sqrt2} usuwamy niewymierność z mianownika
x=\frac{5}{\sqrt2}\cdot \frac{\sqrt2}{\sqrt2}
x=\frac{5\sqrt2}{(\sqrt2)^2}
x=\frac{5\sqrt2}{2}
b)
1 + √3x = x + 2
\sqrt3x-x=2-1
x(\sqrt3-1)=1
x=\frac{1}{\sqrt3-1}=\frac{1\cdot (\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
x=\frac{\sqrt3+1}{(\sqrt3)^2-1^2} wzór skróconego mnożenia a^2-b^2=(a-b)(a+b)
x=\frac{\sqrt3+1}{2}
c)
2 - √2x = x - 4
-\sqrt2x-x=-4-2
-x(\sqrt2+1)=-6 \ |*(-1)
x(\sqrt2+1)=6
x=\frac{6}{\sqrt2+1)}=\frac{6\cdot (\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}
x=\frac{6(\sqrt2-1)}{2-1}
x=6(\sqrt2-1)
d)
2x + 3 + x√3 = 4
2x+x\sqrt3=4-3
x(2+\sqrt3)=1
x=\frac{1}{2+\sqrt3}-\frac{1\cdot (2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)}
x=\frac{2-\sqrt3}{4-3}
x=2-\sqrt3
x=
e)
(x + 2√2) √2 = √3x - 1
\sqrt2x+2\cdot 2=\sqrt3x-1
\sqrt2x-\sqrt3x=-1-4
x(\sqrt2-\sqrt3)=-5
x=\frac{-5}{\sqrt2-\sqrt3}=\frac{-5\cdot (\sqrt2+\sqrt3)}{(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}
x=\frac{-5\sqrt2+\sqrt3)}{2-3}
x=\frac{-5\sqrt2+\sqrt3)}{-1}
x=5(\sqrt2+\sqrt3)
f)
√5(x - √5) = 10 - x
\sqrt5(x-\sqrt5)=10-x
\sqrt5x-5=10-x
\sqrt5x+x=10+5
x(\sqrt5+1)=15
x=\frac{15}{\sqrt5+1}=\frac{15\cdot (\sqrt5-1)}{(\sqrt5+1)(\sqrt5-1)}
x=\frac{15(\sqrt5-1)}{4}