Răspuns :
[tex]\text{Al k+1 termen dintr-o dezvoltare este :} \boxed{T_{k+1} =C_n^k \cdot a^{n-k} \cdot b^k}\\
\text{Din diferenta coeficientilor binomiali il putem afla pe n:}\\
C_n^2 -C_n^1 =27\\
\dfrac{n(n-1)}{2}-n =27\\
n^2-n-2n=54\\
n^2-3n-54=0\\
\Delta =9+4\cdot 54= 9+216 =225 \Rightarrow \sqrt{\Delta} =15\\
n_1= \dfrac{3+15}{2}=9\\
n_2=\dfrac{3-15}{2}= -6\notin\mathbb{N}\text{,deci solutia aceasta nu convine}\\
\text{Asadar}\ T_{k+1} =C_{9}^k\cdot \left(\dfrac{1}{\sqrt[8]{x}}\right)^{9-k}\cdot \left(x^{\lg x}\right)^k[/tex]
[tex]T_2=C_9^1\cdot \left(\dfrac{1}{\sqrt[8]{x}}\right)^8 \cdot \left (x^{\lg x}\right)=9\cdot \dfrac{1}{x}\cdot x^{\lg x}= 9 \cdot x^{lg x-1} \\
T_2=900\\
9\cdot x^{lg x-1} =900\\
x^{\lg x-1 }=100 |\lg()\\
(\lg x-1)\cdot \lg x=2\\
\text{Notam } \lg x=t\\
(t-1)\cdot t=2\\
t^2-t-2=0\\
(t-2)(t+1)=0\Rightarrow t\in \{2,-1\}\\
i)\text{Pentru }t=2\Rightarrow x=100[/tex]
[tex]ii)\text{Pentru } t=-1 \Rightarrow x=\dfrac{1}{10}\\
S:x\in \left\{100,\dfrac{1}{10}\right \} [/tex]
