Question

Am nevoie de ajutor la acest exercitiu! Multumesc!

Answer 1

Răspuns:

x=2

Explicație pas cu pas:

$C_{n} ^{n}+C_{n} ^{n-1}+C_{n} ^{n-2}=22, \frac{n!}{n!*0!}+\frac{n!}{(n-1)!*1!}+\frac{n!}{(n-2)!2!}=22\\1+n+\frac{n(n-1)}{2}=22, 2+2n+n(n-1)=44, n^{2}+n=42, n(n+1)=6*7, deci n=6$

$T_{3}+T_{5}=135$

$T_{3}=T_{2+1}=C_{6}^{2}*(\sqrt{2^{x} })^{4}*(\sqrt{2^{1-x} })^{2} =15*2^{x+1}\\T_{5}=T_{4+1}=C_{6}^{4}*(\sqrt{2^{x} })^{2}*(\sqrt{2^{1-x} })^{4} =15*2^{2-x}\\Deci 15*2^{x+1}+15*2^{2-x}=135 , 2^{x+1}+2^{2-x}=9, 2*2^{x}+\frac{4}{2^{x}} =9, 2*(2^{x})^{2}-9*2^{x}+4=0\\ x=-1 sau x=2$

Answer 2

Răspuns:

Explicație pas cu pas:

$C_{n} ^{n-2}+C_{n} ^{n-1}+C_{n} ^{n}=22 \\\frac{n(n-1)}{2}+n+1=22\\n^{2} +n-42=0$

cu n = 6

$T_{3}+T_{5} =C_{6} ^{2}(\sqrt{2^{x}} )^{4}(\sqrt{2^{1-x}} )^{2} +C_{6} ^{4}(\sqrt{2^{x}} )^{2}(\sqrt{2^{1-x}} )^{4}=\\=15*2^{2x}*2^{1-x}+15*2^{x}*2^{2-2x}=\\=15*2^{x}*2^{1-x}(2^{x}+2^{1-x})=\\=15*2^{x}\frac{2}{2^{x}} (2^{x}+\frac{2}{2^{x}}) =\\=30(2^{x}+\frac{2}{2^{x}}) \\30(2^{x}+\frac{2}{2^{x}}) =135\\2(2^{x}+\frac{2}{2^{x}}) =9$

Notam $2^{x}=t$

$2(t+\frac{2}{t})=9\\ 2t^{2} -9t+4=0$

Δ=49 ⇒$t_{1}=4$ si $t_{2}=\frac{1}{2}$

$2^{x}=4$⇒ x=2

$2^{x}=\frac{1}{2}$ ⇒ x=-1

Deci varianta B, x = 2