Ответ равен 7,3 это мое предположение
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![1)\frac{m^{2}-k^{2}}{mk+m^{2}}:\frac{m^{2}-2mk+k^{2}}{k^{2}}=\frac{(m-k)(m+k)}{m(k+m)}*\frac{k^{2}}{(m-k)^{2}}=\frac{k^{2}}{m(m-k)}\\\\\frac{k}{k-m}+\frac{k^{2}}{m(m-k)}=\frac{k}{k-m}-\frac{k^{2}}{m(k-m)}=\frac{mk-k^{2}}{m(k-m)}=\frac{k(m-k)}{m(k-m)}=-\frac{k}{m}\\\\-\frac{k}{m}=-\frac{k}{m}](https://tex.z-dn.net/?f=1%29%5Cfrac%7Bm%5E%7B2%7D-k%5E%7B2%7D%7D%7Bmk%2Bm%5E%7B2%7D%7D%3A%5Cfrac%7Bm%5E%7B2%7D-2mk%2Bk%5E%7B2%7D%7D%7Bk%5E%7B2%7D%7D%3D%5Cfrac%7B%28m-k%29%28m%2Bk%29%7D%7Bm%28k%2Bm%29%7D%2A%5Cfrac%7Bk%5E%7B2%7D%7D%7B%28m-k%29%5E%7B2%7D%7D%3D%5Cfrac%7Bk%5E%7B2%7D%7D%7Bm%28m-k%29%7D%5C%5C%5C%5C%5Cfrac%7Bk%7D%7Bk-m%7D%2B%5Cfrac%7Bk%5E%7B2%7D%7D%7Bm%28m-k%29%7D%3D%5Cfrac%7Bk%7D%7Bk-m%7D-%5Cfrac%7Bk%5E%7B2%7D%7D%7Bm%28k-m%29%7D%3D%5Cfrac%7Bmk-k%5E%7B2%7D%7D%7Bm%28k-m%29%7D%3D%5Cfrac%7Bk%28m-k%29%7D%7Bm%28k-m%29%7D%3D-%5Cfrac%7Bk%7D%7Bm%7D%5C%5C%5C%5C-%5Cfrac%7Bk%7D%7Bm%7D%3D-%5Cfrac%7Bk%7D%7Bm%7D)
Что и требовалось доказать
(10^9-95)
Число 10 в 9-ой степени - 1000000000 (10*10*10*10*10*10*10*10*10=1000000000)
1000000000-95=999999906.
9+9+9+9+9+9+9+0+6=69.
Ответ: 69.
Решение
2lg5 + 1/2lg16 = lg(5^2) + lg√(16) = lg(25*4) = lg100 = lg(10)^2 = 2lg10 = 2