[toán 8]Phương trình dạng phân số

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01263812493

Giải phương trình sau:
[TEX] \frac{1}{1.2.3}+\frac{1}{2.3.4}+....+ \frac{1}{x(x+1)(x+2)} =\frac{637}{2550}[/TEX]

\Rightarrow [TEX] \frac{2}{1.2.3}+\frac{2}{2.3.4}+....+ \frac{2}{x(x+1)(x+2)} =\frac{637}{1275} mà \frac{2}{1.2.3}= \frac{1}{1.2} - \frac{1}{2.3} \frac{1}{2.3.4}= \frac{1}{2.3} - \frac{1}{3.4}[/TEX]
cứ thế mà làm ra :D
 
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0

01263812493

[TEX]\Rightarrow 2(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{x(x+1)(x+2)})=\frac{637}{1275}\\\Leftrightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{x(x+1)(x+2)}=\frac{637}{1275}\\\Leftrightarrow \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+......+\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}=\frac{637}{1275}\\\Leftrightarrow \frac{1}{1.2}-\frac{1}{(x+1)(x+2)}=\frac{637}{1275}\\\Leftrightarrow\frac{(x+1)(x+2)-2}{2(x+1)(x+2)}=\frac{637}{1275}[/TEX] từ doa' giải pt
 
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