[toán11] Giải các phương trình

I

i_am_shy

2)
Đk: 3 \leq x $\in$ N.

BPT \Leftrightarrow $\frac{1}{2}.\frac{(2x)!}{(2x - 2)!} - \frac{x!}{(x - 2)!}$ \leq $\frac{6}{x}.\frac{x!}{3!(x - 3)!} + 10$


\Leftrightarrow $\frac{1}{2}.(2x - 1)2x - (x-1)x$ \leq $\frac{6}{x}. \frac{(x-2)(x-1)x}{3!} + 10$

\Leftrightarrow $(2x - 1)x - (x -1)x$ \leq $(x-2)(x-1) +10$

\Leftrightarrow $3x - 12 $\leq 0

\Leftrightarrowx \leq 4

=> No là : 3 \leq x \leq 4.

P/s: Cậu này hay nhỉ, ;))
 
H

huytrandinh

[TEX]P_{x}.A_{x}^{2}=x!.x(x-1)[/TEX]
[TEX]pt<=>x!.x.(x-1)+72=12.x!+6.x.(x-1)[/TEX]
[TEX]<=>x!(x^{2}-x-12)=6(x^{2}-x-12)[/TEX]
[TEX].x^{2}-x-12=0=>x=4[/TEX]
[TEX].x^{2}-x-12\neq 0=>x!=6<=>x=3[/TEX]
 
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Z

zkyske

[TEX]1)P_x.A^2_x+72=6.(A^2_x+2P_x)[/TEX]
\Leftrightarrow[TEX]x!.x(x-1) + 72 - 6x(x-1)-12x! = 0[/TEX]
\Leftrightarrow[TEX] x(x-1)(x!-6x)+12(x!-6x)=0[/TEX]
\Leftrightarrow[TEX](x!-6x)[x(x-1)+12]=0[/TEX]
\Leftrightarrow[TEX]x!-6=0 v x^2-x+12=0[/TEX]
\Leftrightarrow[TEX]x= 3 v VN[/TEX]
 
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