Задача 63289 5 и 6 задача...
Условие
Решение
Все решения
[m]log_{2}14=log_{2}2\cdot 7=log_{2}2+log_{2}7=1+log_{2}7[/m]
[m]log_{2}56=log_{2}8\cdot 7=log_{2}8+log_{2}7=3+log_{2}7[/m]
Тогда
[m]\frac{log^2_{2}14+(log_{2}14)\cdot( log_{2}56)-2log^2_{2}56 }{log_{2}14-log_{2}56}=\frac{(1+log_{2}7)^2+(1+log_{2}7)\cdot (3+log_{2}7)-2\cdot (3+log_{2}7)^2 }{1+log_{2}7-(3+log_{2}7)}=\frac{1+2log_{2}7+log^2_{2}7+3+4log_{2}7+log^2_{2}7-18-12log_{2}7-2log^2_{2}7}{1-3}=3log_{2}7+7[/m]
О т в е т. [m]3\cdot log_{2}7+7[/m]
6.
[m]log_{5}7\sqrt{5}=log_{5}7+log_{5}\sqrt{5}=log_{5}7+\frac{1}{2}log_{5}5=log_{5}7+\frac{1}{2}[/m]
[m]log_{5}49=log_{5}7^2=2log_{5}7[/m]
Тогда
[m]\frac{log^2_{5}7\sqrt{5}+2log^2_{5}7-3\cdot (log_{5}7\sqrt{5})\cdot( log_{5}7)}{log_{5}7\sqrt{5}-log_{5}49}=\frac{(log_{5}7+\frac{1}{2})^2+2log^2_{5}7-3\cdot (log_{5}7+\frac{1}{2})\cdot( log_{5}7)}{log_{5}7+\frac{1}{2}-2log_{5}7}=\frac{log^2_{5}7+log_{5}7+\frac{1}{4}+2log^2_{5}7-3\cdot log^2_{5}7-\frac{3}{2}log_{5}7}{\frac{1}{2}-log_{5}7}=\frac{\frac{1}{4}-\frac{1}{2}log_{5}7}{\frac{1}{2}-log_{5}7}=\frac{\frac{1}{2}\cdot (\frac{1}{2}-log_{5}7)}{\frac{1}{2}-log_{5}7}=\frac{1}{2}[/m]
О т в е т. [m]\frac{1}{2}[/m]