Giải PT sau:
$\sqrt{x+1}-\sqrt{4-x}+8-x^2=0$
GPT: $\sqrt{x+1}-\sqrt{4-x}+8-x^2=0$
Started By mango, 08-01-2013 - 08:40
#1
Posted 08-01-2013 - 08:40
#2
Posted 08-01-2013 - 10:40
ĐKXĐ:$-1\leq x\leq4$Giải PT sau:
$\sqrt{x+1}-\sqrt{4-x}+8-x^2=0$
Ta có $PT \Leftrightarrow \sqrt{x+1}-2-\sqrt{4-x}+1+8-x^2+1=0$
$\Leftrightarrow (x-3).(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{4-x}+1}-x-3)=0$
Và khi đó thì $x=3$ , do điều kiện ban đầu nên $\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{4-x}+1}-x-3$ vô nghiệm
Vậy $x={3}$
Edited by thanhdotk14, 08-01-2013 - 10:45.
- N H Tu prince and mango like this
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