1)objętość prostopadłościanu wynosi 125cm³ a jego wysokość będąca najdłuższą krawędzią jest równa 10cm. Wyznacz długość i szerokość podstawy prostopadłościanu, wiedząc że wymiary tej bryły tworzą ciąg geometryczny
2)udowidnij ze nierownosc a²+b²+c²≥ab+bc+ac jest prawdziwa dla dowolnych liczb rzeczywistych a,b i c.
;)
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1)
-krawędź podstawy
-krawędz podstawy
-wysokość
![V=P_ph V=P_ph](https://tex.z-dn.net/?f=V%3DP_ph)
![125=P_p\cdot 10 125=P_p\cdot 10](https://tex.z-dn.net/?f=125%3DP_p%5Ccdot+10)
![P_p=12,5 P_p=12,5](https://tex.z-dn.net/?f=P_p%3D12%2C5)
![\begin{cases}a \cdot aq=12,5\\ aq^2=10\end{cases} \begin{cases}a \cdot aq=12,5\\ aq^2=10\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da+%5Ccdot+aq%3D12%2C5%5C%5C+aq%5E2%3D10%5Cend%7Bcases%7D)
![\begin{cases}a^2q=12,5\\ a=\frac{10}{q^2}\end{cases} \begin{cases}a^2q=12,5\\ a=\frac{10}{q^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da%5E2q%3D12%2C5%5C%5C+a%3D%5Cfrac%7B10%7D%7Bq%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}(\frac{10}{q^2})^2q=12,5\\ a=\frac{10}{q^2}\end{cases} \begin{cases}(\frac{10}{q^2})^2q=12,5\\ a=\frac{10}{q^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%28%5Cfrac%7B10%7D%7Bq%5E2%7D%29%5E2q%3D12%2C5%5C%5C+a%3D%5Cfrac%7B10%7D%7Bq%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}\frac{100}{q^4}q=12,5\\ a=\frac{10}{q^2}\end{cases} \begin{cases}\frac{100}{q^4}q=12,5\\ a=\frac{10}{q^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cfrac%7B100%7D%7Bq%5E4%7Dq%3D12%2C5%5C%5C+a%3D%5Cfrac%7B10%7D%7Bq%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}\frac{100}{q^3}=12,5\\ a=\frac{10}{q^2}\end{cases} \begin{cases}\frac{100}{q^3}=12,5\\ a=\frac{10}{q^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cfrac%7B100%7D%7Bq%5E3%7D%3D12%2C5%5C%5C+a%3D%5Cfrac%7B10%7D%7Bq%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}q^3=8\\ a=\frac{10}{q^2}\end{cases} \begin{cases}q^3=8\\ a=\frac{10}{q^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dq%5E3%3D8%5C%5C+a%3D%5Cfrac%7B10%7D%7Bq%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}q=2\\ a=\frac{10}{2^2}\end{cases} \begin{cases}q=2\\ a=\frac{10}{2^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dq%3D2%5C%5C+a%3D%5Cfrac%7B10%7D%7B2%5E2%7D%5Cend%7Bcases%7D)
![\begin{cases}q=2\\ a=\frac{10}{4}\end{cases} \begin{cases}q=2\\ a=\frac{10}{4}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dq%3D2%5C%5C+a%3D%5Cfrac%7B10%7D%7B4%7D%5Cend%7Bcases%7D)
![\begin{cases}q=2\\ a=2,5\end{cases} \begin{cases}q=2\\ a=2,5\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dq%3D2%5C%5C+a%3D2%2C5%5Cend%7Bcases%7D)
![b=aq=2,5\cdot 2=5 b=aq=2,5\cdot 2=5](https://tex.z-dn.net/?f=b%3Daq%3D2%2C5%5Ccdot+2%3D5)
![a^2+b^2+c^2 \ge ab+bc+ac a^2+b^2+c^2 \ge ab+bc+ac](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%2Bc%5E2+%5Cge+ab%2Bbc%2Bac)
![a^2-ab+b^2-bc+c^2-ac\ge 0\ / \cdot 2 a^2-ab+b^2-bc+c^2-ac\ge 0\ / \cdot 2](https://tex.z-dn.net/?f=a%5E2-ab%2Bb%5E2-bc%2Bc%5E2-ac%5Cge+0%5C+%2F+%5Ccdot+2)
![2a^2-2ab+2b^2-2bc+2c^2-2ac\ge 0 2a^2-2ab+2b^2-2bc+2c^2-2ac\ge 0](https://tex.z-dn.net/?f=2a%5E2-2ab%2B2b%5E2-2bc%2B2c%5E2-2ac%5Cge+0)
![a^2-2ab+b^2 + b^2-2bc+c^2 + c^2-2ca+a^2\ge 0 a^2-2ab+b^2 + b^2-2bc+c^2 + c^2-2ca+a^2\ge 0](https://tex.z-dn.net/?f=a%5E2-2ab%2Bb%5E2+%2B+b%5E2-2bc%2Bc%5E2+%2B+c%5E2-2ca%2Ba%5E2%5Cge+0)
![(a-b)^2+(b-c)^2+(c-a)^2\ge 0 (a-b)^2+(b-c)^2+(c-a)^2\ge 0](https://tex.z-dn.net/?f=%28a-b%29%5E2%2B%28b-c%29%5E2%2B%28c-a%29%5E2%5Cge+0)
Obliczam pole podsatwy
Obliczam krawędzie podstawy
Ponieważ wysokość była najdłuższym bokiem więc q>0
2)
Suma liczb nieujemnych jest liczbą nieujemną