Răspuns:
Explicație pas cu pas:
[tex]C_{n}^{1}+C_{n}^{3}=2*C_{n}^{2},~\frac{n!}{(n-1)!}+\frac{n!}{3!*(n-3)!}=2*\frac{n!}{2!*(n-2)!}\\n+\frac{n(n-1)(n-2)}{6}=2*\frac{n(n-1)}{2},~|*\frac{6}{n}~6+(n-1)(n-2)=6(n-1),~\\n^{2}-9n+14=0,~delta=25,~n_{1} =2,~n_{2} =7.~dar~n>=4,~deci~n=7.\\a^{n-k}*b^{k}=(\sqrt[3]{\frac{x}{\sqrt{y}}} )^{7-k}*(\sqrt{\frac{\sqrt[3]{y}}{x}})^{k}=\frac{x^{\frac{7-k}{3}}}{y^{\frac{7-k}{6}}}*\frac{y^{\frac{k}{6} }}{x^{\frac{k}{2}}}=x^{\frac{7-k}{3}-\frac{k}{2}}*y^{\frac{k}{6}-\frac{7-k}{6}},~deci~\\[/tex]
[tex]\frac{7-k}{3}-\frac{k}{2}=\frac{k}{6}-\frac{7-k}{6},~|*6\\2(7-k)-3k=k-7+k,~7k=21,~k=3.\\Deci~termenul~este~T_{k+1}=T_{3+1}=T_{4}\\[/tex]
La ex4. introduci în ecuaţie înloc de x variantele de răspuns şi primeşti egalităţi adevărate numai pentru valorile din subpunctul A.
[tex]pt.~x=1,~\sqrt{x-1}+\sqrt[3]{2-x}=\sqrt{1-1}+\sqrt[3]{2-1}=0+1=1,~adevarat \\pt.~x=2,~\sqrt{x-1}+\sqrt[3]{2-x}=\sqrt{2-1}+\sqrt[3]{2-2}=1+0=1,~adevarat \\pt.~x=10,~\sqrt{x-1}+\sqrt[3]{2-x}=\sqrt{10-1}+\sqrt[3]{2-10}=3+(-2)=1,~adevarat \\[/tex]
