Zadanie 1.10.
Oblicz x
a)
x\cdot (\sqrt[4]3+\sqrt[4]2)(\sqrt[4]3-\sqrt[4]2)=\sqrt{48}-\sqrt{32}
x\cdot (\sqrt[4]3)^2-(\sqrt[4]2)^2=\sqrt{16*3}-\sqrt{16*2}
x\cdot (3^{\frac{1}{4}})^2-(2^{\frac{1}{4}})^2=4\sqrt3-4\sqrt2
x\cdot 3^{\frac{1}{2}}-2^{\frac{1}{2}}=4(\sqrt3-\sqrt2)
x\cdot (\sqrt3-\sqrt2)=4\sqrt3-\sqrt2) \ |:(\sqrt3-\sqrt2)
x=4
b)
4^{-0,25}-2^{0,5}=\frac{x}{4^{-0,25}+(2\sqrt2)^{\frac{1}{3}}}
(2^2)^{-0,25}-\sqrt2=\frac{x}{(2^2)^{-0,25}+(2^{\frac{3}{2}})^{\frac{1}{3}}}
2^{-0,5}-2^{0,5}=\frac{x}{2^{-0,5}+2^{\frac{1}{2}}}
\frac{2^{-0,5}-2^{0,5}}{1}=\frac{x}{2^{-0,5}+2^{0,5}} mnożę “na krzyż”
x=(2^{-0,5}-2^{0,5})(2^{-0,5}+2^{0,5})
x=(2^{-0,5})^2-(2^{0,5})^2
x=2^{-1}-2^1
x=\frac{1}{2}-2
x=-1\frac{1}{2}
c)
\frac{\sqrt[3]{25}\cdot 5^{-\frac{1}{2}}}{(\sqrt[6]{25})^2\cdot \sqrt5\cdot x}=(\frac{1}{5})^2\cdot (\frac{1}{\sqrt[4]{25}})^2
\frac{(5^2)^{\frac{1}{3}}\cdot 5^{-\frac{1}{2}}}{(25^{\frac{1}{6}})^2\cdot \sqrt5x}=5^{-2}\cdot (\sqrt[4]{5^2})^{-2}
\frac{5^{\frac{4}{6}}\cdot 5^{-\frac{3}{6}}}{(5^2)^{\frac{1}{3}}\cdot 5^{\frac{1}{2}}\cdot x}=5^{-2}\cdot (5^{\frac{2}{4}})^{-2}
\frac{5^{\frac{1}{6}}}{5^{\frac{2}{3}+\frac{1}{2}}\cdot x}=5^{-2}\cdot 5^{-1}
\frac{5^{\frac{1}{6}}}{5^{\frac{4}{6}+\frac{3}{6}}\cdot x}=5^{-3}
\frac{5^{\frac{1}{6}-\frac{7}{6}}}{x}=(\frac{1}{5})^3
\frac{5^{-1}}{x}=\frac{1}{5^3}
\frac{1}{5x}=\frac{1}{125} mnoże “na krzyż”
5x=125 \ |:5
x=25
d)
\frac{x+5,5}{14}(4+\sqrt2)=\frac{(\sqrt[3]3-\sqrt[3]2)(\sqrt[3]9+\sqrt[3]6+\sqrt[3]4}{8^{\frac{2}{3}}-2^{\frac{1}{2}}}
\frac{(x+5,5)(4+\sqrt2)}{14}=\frac{\sqrt[3]{27}+\sqrt[3]{18}+\sqrt[3]{12}-\sqrt[3]{18}-\sqrt[3]{12}-\sqrt[3]{8}}{(2^3)^{\frac{2}{3}}-\sqrt2}
\frac{(x+5,5)(4+\sqrt2)}{14}=\frac{3-2}{2^2-\sqrt2} mnożę “na krzyż”
(x+5,5)(4+\sqrt2)\cdot (4-\sqrt2)=1\cdot 14
(x+5,5)(16-2)=14
(x+5,5)\cdot 14=14 \ |:14
x=1-5,5
x=-4,5