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Problem #806
giải pt
Status: | New | Start Date: | 26-11-2010 | |
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Priority: | Normal | Due date: | ||
Assigned to: | - | % Done: |
0% |
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Category: | - | |||
Target version: | - | |||
Votes: | 0/0 |
Description
a/ (3x/x^2-3x+1)+(7x/x^2+x+1)=-4
b/ (x^2+x-5/x)+4=(-3x/x^2+x-5)
History
Updated by tanphu over 2 years ago
a)
\(\dfrac{3x}{x^2-3x+1}+\dfrac{7x}{x^2+x+1}=-4\)
\(\Leftrightarrow \dfrac{3}{x-3+\dfrac{1}{x}}+\dfrac{7}{x+1+\dfrac{1}{x}}\)
Đặt \(t=x+\dfrac{1}{x}\) rồi giải tiếp
b)
\(\dfrac{x^2+x-5}{x}+4=\dfrac{-3x}{x^2+x-5}\)
\(\Leftrightarrow x+1-\dfrac{5}{x}+4=\dfrac{-3}{x+1-\dfrac{5}{x}}\)
Đặt \(t=x-\dfrac{5}{x}\) rồi giải tiếp.