[Toán 8]Giải phương trình

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link.123

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tiendat102

Bài 1: [TEX]=(\frac{29-x}{21}+1)+(\frac{27-x}{23}+1)+(\frac{25-x}{25}+1)+(\frac{23-x}{27}+1)+(\frac{21-x}{29}+1)=0[/TEX]

[TEX]=\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0[/TEX]

[TEX]=(50-x)(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29})=0[/TEX]
\Leftrightarrow [TEX]50-x=0[/TEX]
\Leftrightarrow [TEX]x=50[/TEX]
 
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teucon

Bài 1: [TEX]=(\frac{29-x}{21}+1)+(\frac{27-x}{23}+1)+(\frac{25-x}{25}+1)+(\frac{23-x}{27}+1)+(\frac{21-x}{29}+1)=0[/TEX]

[TEX]=\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0[/TEX]

[TEX]=(50-x)(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29})=0[/TEX]
\Leftrightarrow [TEX]50-x=0[/TEX]
\Leftrightarrow [TEX]x=50[/TEX]
Bài này bạn cần phải giải thích[TEX] (\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29})[/TEX] khác 0 nữa. Như vậy mới chặt chẽ
 
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