Bài tập nâng cao · Bài 62
Bài tập nâng cao Hoán vị, chỉnh hợp và tổ hợp · Bài 62
Giải bất phương trình sau:
a. $2C_{x+1}^2+3A_x^2<30$
b. $A_x^3+15<15x$
c. $C_{n+1}^{n-2}-C_{n+1}^{n-1}\le100$.
a. $2C_{x+1}^2+3A_x^2<30$
b. $A_x^3+15<15x$
c. $C_{n+1}^{n-2}-C_{n+1}^{n-1}\le100$.
Xem lời giải
Lời giải
a. $2C_{x+1}^2+3A_x^2<30$
$\Leftrightarrow\dfrac{2(x+1)!}{(x-1)!2!}+\dfrac{3x!}{(x-2)!}<30\Leftrightarrow x(x+1)+3x(x-1)<30$
$\Leftrightarrow4x^2-2x-30<0\Leftrightarrow x\in\left(-\dfrac52;3\right)\Rightarrow x=2$.
b. $A_x^3+15<15x$
$\Leftrightarrow\dfrac{x!}{(x-3)!}+15<15x\Leftrightarrow x(x-1)(x-2)+15-15x<0$
$\Leftrightarrow x^3-3x^2-13x+15<0\Leftrightarrow(x-1)(x+3)(x-5)<0$
$\Leftrightarrow x\in(-\infty;-3)\cup(1;5)\Rightarrow x\in\{3;4\}$.
c. $C_{n+1}^{n-2}-C_{n+1}^{n-1}\le100$
$\Leftrightarrow\dfrac{(n+1)!}{(n-2)!3!}-\dfrac{(n+1)!}{(n-1)!2!}\le100$
$\Leftrightarrow\dfrac{(n+1)n(n-1)}6-\dfrac{(n+1)n}2\le100\Leftrightarrow n^3-3n^2-4n-600\le0$
$\Rightarrow n\in\{2;3;4;5;6;7;8;9\}$.
$\Leftrightarrow\dfrac{2(x+1)!}{(x-1)!2!}+\dfrac{3x!}{(x-2)!}<30\Leftrightarrow x(x+1)+3x(x-1)<30$
$\Leftrightarrow4x^2-2x-30<0\Leftrightarrow x\in\left(-\dfrac52;3\right)\Rightarrow x=2$.
b. $A_x^3+15<15x$
$\Leftrightarrow\dfrac{x!}{(x-3)!}+15<15x\Leftrightarrow x(x-1)(x-2)+15-15x<0$
$\Leftrightarrow x^3-3x^2-13x+15<0\Leftrightarrow(x-1)(x+3)(x-5)<0$
$\Leftrightarrow x\in(-\infty;-3)\cup(1;5)\Rightarrow x\in\{3;4\}$.
c. $C_{n+1}^{n-2}-C_{n+1}^{n-1}\le100$
$\Leftrightarrow\dfrac{(n+1)!}{(n-2)!3!}-\dfrac{(n+1)!}{(n-1)!2!}\le100$
$\Leftrightarrow\dfrac{(n+1)n(n-1)}6-\dfrac{(n+1)n}2\le100\Leftrightarrow n^3-3n^2-4n-600\le0$
$\Rightarrow n\in\{2;3;4;5;6;7;8;9\}$.