Szczegółowe wyjaśnienie:
[tex]Zad.1\\\\x=2-\sqrt3\\\\x^2=(2-\sqrt3)^2=2^2-2\cdot2\cdot\sqrt3+(\sqrt3)^2=4-4\sqrt3+3=7-4\sqrt3\\\huge\boxed{A.\ 7-4\sqrt3}\\\\wzor:\ (a-b)^2=a^2-2ab+b^2[/tex]
[tex]Zad.2\\(a+2\sqrt3)^2=13+4\sqrt3\\a^2+2\cdot a\cdot2\sqrt3+(2\sqrt3)^2=13+4\sqrt3\\a^2+4a\sqrt3+12=13+4\sqrt3\\(a^2+12)+(4a\sqrt3)=13+4\sqrt3\Rightarrow a^2+12=12\ \wedge\ 4a\sqrt3=4\sqrt3\\\\\huge\boxed{B.\ a=1}[/tex]
[tex]Zad.3\\\\a)\ \dfrac{4^5\cdot5^4}{20^4}=\dfrac{4^5\cdot5^4}{(4\cdot5)^4}=\dfrac{4^5\cdot5^4}{4^4\cdot5^4}=4^{5-4}\cdot5^{4-4}=4\\\\b)\ 5^8\cdot16^{-2}=5^8\cdot\left(\dfrac{1}{16}\right)^2=5^8\cdot\left(\dfrac{1}{2^4}\right)^2=5^8\cdot\dfrac{1}{(2^4)^2}=5^8\cdot\dfrac{1}{2^8}=\left(\dfrac{5}{2}\right)^8\\\\c)\ \sqrt[3]{54}-\sqrt[3]2=\sqrt[3]{2\cdot27}-\sqrt[3]2=\sqrt[3]2\cdot\sqrt[3]{27}-\sqrt[3]2=\sqrt[3]2\cdot(\sqrt[3]{27}-1})\\=\sqrt[3]2\cdot(3-1)=2\sqrt[3]2[/tex]
[tex]d)\ (2\sqrt7-5)^2\cdot(2\sqrt7+5)^2=\bigg[(2\sqrt7-5)(2\sqrt7+5)\bigg]^2=\bigg[(2\sqrt7)^2-5^2\bigg]^2\\=\left[2^2(\sqrt7)^2-25\right]^2=\left(4\cdot7-25\right)^2=(28-25)^2=3^2=9\\\\e)\ \sqrt[3]{\dfrac{7}{3}}\cdot\sqrt[3]{\dfrac{81}{56}}=\sqrt[3]{\dfrac{7\!\!\!\!\diagup^1}{3\!\!\!\!\diagup_1}\cdot\dfrac{81\!\!\!\!\!\diagup^{27}}{56\!\!\!\!\!\diagup_8}}=\sqrt[3]{\dfrac{27}{8}}=\dfrac{\sqrt[3]{27}}{\sqrt[3]8}=\dfrac{3}{2}[/tex]
skorzystałem z twierdzeń:
[tex](a\cdot b)^n=a^n\cdot b^n\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\\sqrt[3]{a\cdot b}=\sqrt[3]{a}\cdot\sqrt[3]{b}\\\\(\sqrt{a})^2=a\\\\\sqrt[3]{\dfrac{a}{b}}=\dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}[/tex]
w b) nie obliczałem do końca. Sądzę, że jest tam chyba błąd drukarski. Powinno być 16².
Wówczas wyszłoby 5⁸ · 2⁸ = (2 · 5)⁸ = 10⁸ = 100 000 000
