Szczegółowe wyjaśnienie:
[tex]1.\ 2^4:2^2-2^0=2^{4-2}-1=2^2-1=4-1=\huge\boxed{3}[/tex]
Skorzystałem z:
[tex]a^n:a^m=a^{n-m},\ a\neq0\\\\a^0=1[/tex]
[tex]2.\ 15^8:15^6=15^{8-6}=\huge\boxed{15^2}[/tex]
Skorzystałem z:
[tex]a^n:a^m=a^{n-m},\ a\neq0[/tex]
[tex]3.\ \sqrt{a}=b\iff b^2=a,\ a,b\geq0\\\\a)\ \sqrt{64}+\sqrt{25}=8+5=\huge\boxed{13}\\\\\sqrt{64}=8\ bo\ 8^2=64,\ \sqrt{25}=5\ bo\ 5^2=25\\\\b)\ \sqrt{16\cdot49\cdot100}=\sqrt{16}\cdot\sqrt{49}\cdot\sqrt{100}=4\cdot7\cdot10=\huge\boxed{280}\\\\\sqrt{16}=4\ bo\ 4^2=16,\ \sqrt{49}=7\ bo\ 7^2=49,\ \sqrt{100}=10\ bo\ 10^2=100[/tex]
skorzystałem z:
[tex]\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b},\ a,b\geq0[/tex]
[tex]c)\ \sqrt{3+4^2+9^2}=\sqrt{3+16+81}=\sqrt{100}=\huge\boxed{10}\\\\\sqrt{100}=10\ bo\ 10^2=100[/tex]
[tex]4.\ \left(5^2\right)^4:\left(5^3\right)^2-5^2=5^{2\cdot4}:5^{3\cdot2}-5^2=5^8:5^6-5^2=5^{8-6}-5^2=5^2-5^2=\huge\boxed{0}[/tex]
skorzystałem z:
[tex]\left(a^n\right)^m=a^{n\cdot m}\\\\a^n:a^m=a^{n-m},\ a\neq0[/tex]
[tex]5.\\A(x_A;\ y_A);\ B(x_B;\ y_B)\\\\S_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)\\\\A(4,\ 7);\ B(8,\ -3)\\\\S_{AB}\left(\dfrac{4+8}{2},\ \dfrac{7+(-3)}{2}\right)\to S_{AB}\left(\dfrac{12}{2},\ \dfrac{4}{2}\right)\to \huge\boxed{S_{AB}(6,\ 2)}[/tex]
