żS-4, od: Szemek, zadanie 4

[Liga]

\(\displaystyle{ \left(\sqrt{2^{x}}+ \frac{1}{\sqrt{2^{x-1}}}\right)^m}\)

Suma współczynników trzech pierwszych wyrazów jest równa 22
\(\displaystyle{ {m \choose 0}+{m \choose 1}+{m \choose 2}=22}\), gdzie \(\displaystyle{ m \geq 2}\)
\(\displaystyle{ 1+m+\frac{m(m-1)}{2}=22}\)
\(\displaystyle{ 2+2m+m(m-1)=44}\)
\(\displaystyle{ 2+2m+m^2-m-44=0}\)
\(\displaystyle{ m^2+m-42=0}\)
\(\displaystyle{ (m+7)(m-6)=0}\)
\(\displaystyle{ (m=-7 \vee m=6) \wedge m\geq 2}\)
\(\displaystyle{ m=6}\)

Suma trzeciego i piątego wyrazu dwumianu \(\displaystyle{ \left(\sqrt{2^{x}}+ \frac{1}{\sqrt{2^{x-1}}}\right)^6}\) jest równa 135
\(\displaystyle{ {6 \choose 2} \cdot (\sqrt{2^{x}})^4 \cdot (\frac{1}{\sqrt{2^{x-1}}})^2 + {6 \choose 4} \cdot (\sqrt{2^{x}})^2 \cdot (\frac{1}{\sqrt{2^{x-1}}})^4 = 135}\)
\(\displaystyle{ {6 \choose 2}={6 \choose 4}}\)
\(\displaystyle{ {6 \choose 2}=\frac{5 \cdot 6}{2}=15}\)
\(\displaystyle{ \bigwedge_{\limits{x \in R}} {2^{x}}>0 \wedge \bigwedge_{\limits{x \in R}} {2^{x-1}}>0}\)
\(\displaystyle{ 15 \cdot 2^{2x} \cdot \frac{1}{2^{x-1}} + 15 \cdot 2^x \cdot \frac{1}{2^{2x-2}}=135}\) |:15
\(\displaystyle{ 2^{2x} \cdot {2^{-x+1}} + 2^x \cdot \frac{1}{2^{2x-2}} = 9}\)
\(\displaystyle{ 2^{x+1}+\frac{1}{2^{x-2}}-9=0}\)
\(\displaystyle{ t=2^{x-2},\hbox{ } t>0}\)
\(\displaystyle{ 8t+\frac{1}{t}-9=0}\)|\(\displaystyle{ \cdot t}\)
\(\displaystyle{ 8t^2-9t+1=0}\)
\(\displaystyle{ \Delta_{t}=(-9)^2-4 \cdot 8 \cdot 1}\)
\(\displaystyle{ \Delta_{t}=81-32}\)
\(\displaystyle{ \Delta_{t}=49 \\ \sqrt{\Delta_{t}}=7}\)
\(\displaystyle{ t=\frac{9-7}{16} \vee t=\frac{9+7}{16}}\)
\(\displaystyle{ (t=\frac{1}{8} \vee t=1) \wedge t>0}\)
\(\displaystyle{ t=\frac{1}{8} t=1}\)
\(\displaystyle{ 2^{x-2}=\frac{1}{8} 2^{x-2}=1}\)
\(\displaystyle{ 2^{x-2}=2^{-3} 2^{x-2}=2^0}\)
z różnowartościowości funkcji wykładniczej
\(\displaystyle{ x-2=-3 x-2=0}\)
\(\displaystyle{ (x=-1 x=2) x R}\)
\(\displaystyle{ x \{ -1, 2 \}}\)

[scyth]

5/5