Đáp án: Không có đáp án
Giải thích các bước giải:
ĐKXĐ: $n\ge 1, n\in N$
Ta có:
$\dfrac{1}{C^1_n}-\dfrac{1}{C^2_{n+2}}=\dfrac{7}{6C^1_{n+4}}$
$\to \dfrac{1}{\dfrac{n!}{1!\cdot (n-1)!}}-\dfrac{1}{\dfrac{(n+2)!}{2!\cdot (n+2-2)!}}=\dfrac{7}{6\cdot \dfrac{(n+4)!}{1!\cdot (n+4-1)!}}$
$\to \dfrac{1}{\dfrac{n!}{(n-1)!}}-\dfrac{1}{\dfrac{(n+2)!}{2\cdot n!}}=\dfrac{7}{6\cdot \dfrac{(n+4)!}{(n+3)!}}$
$\to \dfrac{1}{n}-\dfrac{1}{\dfrac{(n+1)(n+2)}{2}}=\dfrac{7}{6\cdot (n+4)}$
$\to \dfrac{1}{n}-\dfrac{2}{(n+1)(n+2)}=\dfrac{7}{6\cdot (n+4)}$
$\to 6\left(n+1\right)\left(n+2\right)\left(n+4\right)-12n\left(n+4\right)=7n\left(n+1\right)\left(n+2\right)$
$\to 6n^3+30n^2+36n+48=7n^3+21n^2+14n$
$\to n^3-9n^2-22n-48=0$
$\to $Vô nghiệm
