Ответы и объяснения

2013-06-29T09:31:49+00:00

42. 1; 43. 15; 44. 5; 45. 2; 46. 7; 47. 12; 48. 2; 49. 2; 50. 1; 51. 4; 52. 2; 53. 3; 54. 4; 55. 3; 56. 4; 57. 2; 58. 2; 59. 2; 60. 4; 61. 2; 62. 3; 63. 4; 65. 2; 66. 3; 67. 3; 68. 4; 69. 1; 70. 4; 71. 25; 72. 24; 73. log₂40; 74. 8; 75. 5; 76. 26; 77. 20; 78. -72; 79. 24; 80. 15; 81. 2; 82. 4; 83. 3; 84. 1; 85. 2; 86. 3; 87. 2; 88. 4; 89. 2; 90. 3; 91. 4; 92. 1; 93. 4; 94. 1; 95. 3; 96. 1; 97. 4; 98. 1; 99. 1; 100. 1; 101. 4; 102. 2; 103. 0

2013-06-29T09:41:13+00:00

 42. 7^{2,5}=7^{\frac{1}{2}}\cdot 49\cdot x^{0,5}\\ 7^{2,5}=7^{0,5}\cdot 7\cdot 7\cdot x^{0,5}\\ 7^{2,5}=7^{0,5+1+1}\cdot x^{0,5}\\ 7^{2,5}=7^{2,5}\cdot x^{0,5} \ | : 7^{2,5}\\ 1=x^{0,5}=\sqrt{x}\\ 1=x

 

43. (0,001)^{-\frac{1}{3}}+2^{-2}\cdot 64^{-\frac{2}{3}}\cdot 4-8^{-1\frac{1}{3}}+(9^{0})^{2}\cdot 5=\\ =\frac{1}{\sqrt[3]{0,001}}+2^{-2}\cdot (2^{6})^{-\frac{2}{3}}\cdot 2^{2}-8^{-\frac{4}{3}}+1\cdot 5=\\ =\frac{1}{\sqrt[3]{0,001}}+2^{-2}\cdot 2^{-4}\cdot 2^{2}-\frac{1}{\sqrt[3]{8^{4}}}+1\cdot 5=\\ =\frac{1}{\sqrt[3]{0,001}}+2^{-2-4+2}-\frac{1}{\sqrt[3]{(2^{3})^{4}}}+1\cdot 5=\\ =\frac{1}{\sqrt[3]{0,001}}+2^{-4}-\frac{1}{\sqrt[3]{2^{12}}}+1\cdot 5=\\

=\frac{1}{0,1}+\frac{1}{2^{4}}-\frac{1}{2^{\frac{12}{3}}}}+5=10+\frac{1}{16}-\frac{1}{2^{4}}+5=10+\frac{1}{16}-\frac{1}{16}+5=15

 

44. 3^{-4}\cdot 27^{-\frac{2}{3}}\cdot 9-27^{-1\frac{1}{3}}+(8^{0})^{3}\cdot 2+(0,125)^{\frac{2}{3}}=\\ =3^{-4}\cdot (3^{3})^{-\frac{2}{3}}\cdot 3^{2}-\frac{1}{\sqrt[3]{27^{4}}}+1\cdot 2+\sqrt[3]{0,125^{2}}=\\ =3^{-4-2+2}-\frac{1}{\sqrt[3]{(3^{3})^{4}}}+2+\sqrt[3]{(0,5^{3})^{2}}=\\ =3^{-4}-\frac{1}{3^{\frac{12}{3}}}+2+0,5^{\frac{6}{3}}=\frac{1}{3^{4}}-\frac{1}{3^{4}}+2+0,5^{2}=\\ =2+0,25=2,25

 

45. 9^{-\frac{5}{2}}+10\cdot (4^{0})^{5}-(0,25)^{-\frac{3}{2}}-9^{-\frac{3}{2}}\cdot 27\cdot 3^{-5}=\\ =\frac{1}{\sqrt{9^{5}}}+10\cdot 1-\frac{1}{\sqrt{0,25^{3}}}-(3^{2})^{-\frac{3}{2}}\cdot 3^{3}\cdot 3^{-5}=\\ =\frac{1}{\sqrt{(3^{2})^{5}}}+10\cdot 1-\frac{1}{\sqrt{(0,5^{2})^{3}}}-3^{-3+3-5}=\\ =\frac{1}{3^{\frac{10}{2}}}+10-\frac{1}{0,5^{\frac{6}{2}}}-3^-5=\frac{1}{3^{5}}+10-\frac{1}{0,5^{3}}-\frac{1}{3^{5}}=\\ =10-\frac{1}{0,125}=10-8=2

 

46. 9^{-\frac{3}{2}}-(5^{0})^{3}\cdot 3+(0,01)^{-\frac{1}{2}}-9\cdot 3^{-3}\cdot 27^{-\frac{2}{3}}=\\ =\frac{1}{\sqrt{9^{3}}}-1\cdot 3+\frac{1}{\sqrt{0,01}}-3^{2}\cdot 3^{-3}\cdot (3^{3})^{-\frac{2}{3}}=\\ =\frac{1}{3^{3}}-3+\frac{1}{0,1}-3^{2-3-2}=\frac{1}{27}-3+10-3^{-3}=\frac{1}{27}+7-\frac{1}{3^{3}}=\\ =\frac{1}{27}-\frac{1}{27}+7=7

 

47. (0,001)^{-\frac{1}{3}}+27^{-2\frac{1}{3}}+(6^{0})^{5}\cdot 2-3^{-4}\cdot 81^{-\frac{3}{2}}\cdot 27=\\ =\frac{1}{\sqrt[3]{0,001}}+(3^{3})^{-2\frac{1}{3}}+1\cdot 2-3^{-4}\cdot (3^{4})^{-\frac{3}{2}}\cdot 3^{3}=\\ =\frac{1}{0,1}+3^{-7}+2-3^{-4-6+3}=10+\frac{1}{3^{7}}+2-3^{-7}=\\ =12+\frac{1}{3^{7}}-\frac{1}{3^{7}}=12

 

48. 64^{-\frac{5}{6}}-(0,125)^{-\frac{1}{3}}-32\cdot 2^{-4}\cdot 16^{-1\frac{1}{2}}-(3^{0})^{4}\cdot 4=\\ =\frac{1}{\sqrt[6]{64^{5}}}-\frac{1}{\sqrt[3]{0,125}}-2^{5}\cdot 2^{-4}\cdot (2^{4})^{-1\frac{1}{2}}-1\cdot 4=\\ =\frac{1}{\sqrt[6]{(2^6)^{5}}}-\frac{1}{0,5}-2^{5-4-6}-4=\frac{1}{2^{\frac{30}{6}}}-2-2^{-5}-4=\\ =\frac{1}{2^{5}}-2-\frac{1}{2^{5}}-4=-2-4=-6

 

49. 16^{-\frac{5}{4}}-(0,01)^{-\frac{1}{2}}+12\cdot (7^{0})^{3}-16\cdot 2^{-5}\cdot 64^{-\frac{2}{3}}=\\ =(2^{4})^{-\frac{5}{4}}-\frac{1}{\sqrt{0,01}}+12\cdot 1-2^{4}\cdot 2^{-5}\cdot (2^{6})^{-\frac{2}{3}}=\\ =2^{-5}-\frac{1}{0,1}+12-2^{4-5-4}=\frac{1}{2^{5}}-10+12-2^{-5}=\frac{1}{2^{5}}+2-\frac{1}{2^{5}}=2