Proszę o rozwiązanie zad 2, 3 i 4 str 36. INTERESUJĄ MNIE CAŁE OBLICZENIA, A NIE SAME WYNIKI! Daje naj ;))

Zad.26⁻³*6⁵=6²=6*6=364⁻⁸*4⁶=4⁻²=(1/4)²=1/160,5⁻⁴*0,5⁻¹=0,5⁻⁵=2⁵=32(2/3)²*(2/3)⁻⁵=(2/3)⁻³=(3/2)³=27/8=2 3/85⁷:5¹⁰=5⁻³=(1/5)³=1/1257⁻⁴:7⁻⁶=7²=7*7=49(3/4)⁻⁵:(3/4)⁻³=(3/4)⁻²=(4/3)²=16/9=1 7/9(3/8)⁷:(3/8)⁸=(3/8)⁻¹=8/3=2 2/3Zad.3(2⁵)²=2¹⁰=1024(5⁻²)⁻²=5⁴=625((3/2)²)⁻²=(3/2)⁻⁴=(2/3)⁴=16/81((5/7)⁻¹)⁻²=(5/7)²=25/49(3⁶)⁻²*3¹⁰=3⁻¹²*3¹⁰=3²=3*3=9(6⁻²)⁵*6⁸=6⁻¹⁰*6⁸=6⁻²=(1/6)²=1/36(4⁻³)⁻⁴:4⁹=4¹²:4⁹=4³=4*4*4=64(8⁴)⁻²:8⁻¹⁰=8⁻⁸:8⁻¹⁰=8²=8*8=64Zad.42³*4⁶=2³*2¹²=2¹⁵4⁻⁵*8²=2⁻¹⁰*2⁶=2⁻⁴64²:32⁻³=2¹²:2⁻¹⁵=2¹²*2¹⁵=2²⁷(16⁻²:4⁻⁸)*8⁴=(2⁻⁸:2⁻¹⁶)*2¹²=2⁸*2¹²=2²⁰Liczę na najlepszą odpowiedź.

[latex]2.\a) 6^{-3}cdot6^{5} = 6^{2} = 36\\b) 4^{-8}cdot4^{6} = 4^{-2} = (frac{1}{4})^{2} = frac{1}{16}\\c) 0,5^{-4}cdot0,5^{-1} = 0,5^{-5} =(frac{1}{2})^{-5} = 2^{5} = 32\\d) (frac{2}{3})^{2}cdot(frac{2}{3})^{-5} =(frac{2}{3})^{-3} = (frac{3}{2})^{3} = frac{27}{8} = 3frac{3}{8}\\e) 5^{7}:5^{10} = 5^{-3} = (frac{1}{5})^{3} = frac{1}{125}\\f) 7^{-4}:7^{-6} = 7^{-4-(-6)} = 7^{2} = 49[/latex] [latex]g) (frac{3}{4})^{-5}:(frac{3}{4})^{-3}= (frac{3}{4})^{-2} = (frac{4}{3})^{2} = frac{16}{9} = 1frac{7}{9}\\h) (frac{3}{8})^{7}:(frac{3}{8})^{8} = (frac{3}8)^{-1} = frac{8}{3} = 2frac{2}{3}[/latex] [latex]3.\a) (2^{5})^{2} = 2^{10} = 1024\\b) (5^{-2})^{-2} = 5^{4} = 625\\c) ((frac{3}{2})^{2})^{-2} = (frac{3}{2})^{-4} = (frac{2}{3})^{4}= frac{16}{81}\\d) ((-frac{5}{7})^{-1})^{-2}= (-frac{5}{7})^{2} =frac{25}{49}\\e) (3^{6})^{-2} cdot3^{10} = 3^{-12}cdot3^{10} = 3^{-2} = (frac{1}{3})^{2} = frac{1}{9}\\f) (6^{-2})^{5}cdot6^{8} = 6^{-10}cdot6^{8} = 6^{-2} = (frac{1}{6})^{2} = frac{1}{36}}\\g) (4^{-3})^{-4}:4^{9} = 4^{12}:4^{9}= 4^{3} = 64[/latex] [latex]h) (8^{4})^{-2}:8^{-10} = 8^{-8}:8^{-10} = 8^{2} = 64[/latex] [latex]4)\a) 2^{3}cdot4^{6} = 2^{3}cdot(2^{2})^{6} = 2^{3}cdot2^{12} = 2^{15}\\b) 4^{-5}cdot8^{2} = (2^{2})^{-5}cdot(2^{3})^{2} = 2^{-10}cdot2^{6} = 2^{-4}\\c) 64^2}:32^{-3} = (2^{6})^{2}:(2^{5})^{-3} = 2^{12}:2^{-15}= 2^{27}\\d) (16^{-2}:4^{-8}cdot8^4}=((2^{4}^{-2}:(2^{2})^{-8}):(2^{3})^{4} = (2^{-8}:2^{-16})cdot2^{12}=\=2^{8}cdot2^{12} = 2^{20[/latex]