Đáp án: $x=6$
Giải thích các bước giải:
ĐKXĐ: $x\ge 1$
$35C^{x-1}_{2x}=132C^x_{2x-2}$
$\to 35\cdot \dfrac{(2x)!}{(x-1)!\cdot (2x-(x-1))!}=132\cdot \dfrac{(2x-2)!}{x!\cdot (2x-2-x)!}$
$\to 35\cdot \dfrac{(2x)!}{(x-1)!\cdot (x+1)!}=132\cdot \dfrac{(2x-2)!}{x!\cdot (x-2)!}$
$\to 35\cdot \dfrac{(2x)!\cdot x!\cdot (x-2)!}{(2x-2)!\cdot (x-1)!\cdot (x+1)!}=132$
$\to 35\cdot \dfrac{(2x)!}{(2x-2)!}\cdot\dfrac{x!}{(x-1)!}\cdot\dfrac{(x-2)!}{(x+1)!}=132$
$\to 35\cdot (2x\cdot (2x-1))\cdot x\cdot\dfrac{1}{(x+1)\cdot x\cdot (x-1)}=132$
$\to 70\cdot \dfrac{x\cdot (2x-1)}{(x+1)(x-1)}=132$
$\to 70x\left(2x-1\right)=132\left(x+1\right)\left(x-1\right)$
$\to 140x^2-70x=132x^2-132$
$\to 8x^2-70x+132=0$
$\to 2(x-6)(4x-11)=0$
$\to x=6$
