Question

giải PT: x+3C2 - xA2 = x-4

Answer 1

ĐKXĐ: x>=2

\(C^2_{x+3}-A^2_x=x-4\)

=>\(\dfrac{\left(x+3\right)!}{\left(x+3-2\right)!\cdot2!}-\dfrac{x!}{\left(x-2\right)!}=x-4\)

=>\(\dfrac{\left(x+2\right)\left(x+3\right)}{2}-x\left(x-1\right)=x-4\)

=>\(\left(x+2\right)\left(x+3\right)-2x\left(x-1\right)=2x-8\)

=>\(x^2+5x+6-2x^2+2x-2x+8=0\)

=>\(-x^2+5x+14=0\)

=>\(x^2-5x-14=0\)

=>(x-7)(x+2)=0

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Answer 2

Đề là \(C_{x+3}^2-A_x^2=x-4\) đúng ko em?