Toán 8

T

thaolovely1412

[TEX]M= \frac{x^2-2x}{3x+3}: \frac{x^2-4}{3x^2+6x+3}[/TEX]
[TEX]= \frac{x(x-2)}{3(x+1)}.\frac{3(x^2+2x+1)}{(x+2)(x-2)}[/TEX]
[TEX]=\frac{x(x-2)3(x+1)^2}{3(x+1)(x+2)(x-2)}[/TEX]
[TEX]=\frac{x(x+1)}{x-2}[/TEX]
 
T

taylory

Câu b nhé!

[TEX]M= \frac{x^2-2x}{3x+3}: \frac{x^2-4}{3x^2+6x+3}[/TEX]
[TEX]= \frac{x(x-2)}{3(x+1)}.\frac{3(x^2+2x+1)}{(x+2)(x-2)}[/TEX]
[TEX]=\frac{x(x-2)3(x+1)^2}{3(x+1)(x+2)(x-2)}[/TEX]
[TEX]=\frac{x^2+x}{x-2}[/TEX]
[TEX]=(x^2-2x)+(3x-6)+6}{x-2}[/TEX]
[TEX]=x+3-\frac{6}{x-2}[/TEX]
\Rightarrow Để M nguyên thì [TEX]\frac{6}{x-2}[/TEX] nguyên \Rightarrow[TEX] x-2 \in \[/TEX] Ư(6)
\Rightarrow [TEX]x-2 \in \ [/TEX]{-6;-3;-2;-1;1;2;3;6} \Rightarrow [TEX]x \in \ [/TEX]{-4;-1;0;1;3;4;5;8}
 
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