Question

вычислите ............................

Answer 1

Ответ:

$1)\ -\frac{3}{4}$

$2)\ \frac{1}{2}$

$3)\ 1$

$4)\ 36\frac{1}{4}$

Объяснение:

$1)\\\\\frac{3}{log_33}-\frac{2}{log_84}-\frac{1}{log_{27}{81}}=\frac{3}{1}-\frac{2}{\frac{lg4}{lg8}}-\frac{1}{\frac{lg81}{lg27}}=3-\frac{2lg8}{lg4}-\frac{lg27}{lg81}=\\\\3-\frac{2lg2^3}{lg2^2}-\frac{lg3^3}{lg3^4}=3-\frac{2\cdot 3lg2}{2lg2}-\frac{3lg3}{4lg3}=3-3-\frac{3}{4}=-\frac{3}{4}$

$2)\\\\\frac{lg 2+lg3}{lg 3,6+1}=\frac{lg(2\cdot 3)}{lg 3,6+lg10}=\frac{lg6}{lg( 3,6\cdot 10)}=\frac{lg6}{lg36}=\frac{lg6}{lg6^2}=\frac{lg6}{2lg6}=\frac{1}{2}$

$3)\\\\2^{2-log_25}+(\frac{1}{2})^{log_25}=2^{2}\cdot 2^{-log_25}+(2^{-1})^{log_25}=4\cdot 2^{-log_25}+2^{-log_25}=\\\\2^{-log_25}\cdot (4+1)=5\cdot \frac{1}{2^{log_25}}=5\cdot\frac{1}{5}=1$

$4)\\\\3^{2+log_34}+(\frac{1}{3})^{log_34}=3^{2}\cdot 3^{log_34}+\frac{1}{3^{log_34}}=9\cdot 4+\frac{1}{4}=36+\frac{1}{4}=36\frac{1}{4}$

Answer 2

Відповідь:

Пояснення:

#  1 )  . . . = 3/1 - 2 : ( log₂4/log₂8 ) - 1 : ( log₃81/log₃27 ) = 3 - 2 : ( 2/3 ) -

                 - 1 : ( 2/3 ) = 3 - 3 - 3/4 = - 3/4 ;

  2 )  . . . = lg( 2 * 3 )/lg( 3,6 * 10 ) = lg6/lg36 = lg6/lg6² = lg6/( 2lg6 ) = 1/2;

  3 )  . . . = 2² * ( 2^( log₂5 ) )⁻¹ +  ( 2^( log₂5 ) )⁻¹ = 4 * 5⁻¹ + 5⁻¹ = 5/5 = 1 ;

  4 )  . . . = 3² * 3^( log₃4 ) + ( 3^( log₃4 ) )⁻¹ = 9 * 4 + 4⁻¹ = 36 1/4 .