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toiyeulopminh

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D

demon311

1) $5 \ge x \ge 0$

$\sqrt{ x}-1+\sqrt{ 5-x}-2=(x-1)(x-4) \\
\dfrac{ x-1}{\sqrt{ x}+1}-\dfrac{ x-1}{\sqrt{ 5-x}+2}=(x-1)(x-4) \\
(x-1)(\dfrac{ 1}{\sqrt{ x-1}+1}-\dfrac{ 1}{\sqrt{5-x}+2}+4-x)=0$

3) $4 \le x \le 6$

$\sqrt{ x-4}-1+\sqrt{ 6-x}-1=2x^2-13x+15 \\
\dfrac{ x-5}{\sqrt{ x-4}+1}-\dfrac{ x-5}{\sqrt{ 6-x}+1}-(2x-3)(x-5)=0 \\
(x-5)(...)=0$
 
D

dien0709

nhân liên hợp

2) ĐK 1/2<=x<=4[TEX]\Rightarrow\sqrt{2x-1}-1+sqrt{x^2+3}-2=1-x[/TEX]
[TEX]\Rightarrow\frac{2(x-1)}{sqrt{2x-1}+1}+\frac{(x-1)(x+1}{sqrt{x^2+3}+2}-(x-1)=0[/TEX]
 
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pe.chizy08

2, ĐK 1/2 \leq x \leq 4(1)
\Leftrightarrow$\sqrt{2x-1} - 1 + \sqrt{x^2 + 3} - 2 = 1-x$
\Leftrightarrow$\frac{2x-1-1}{\sqrt{2x-1}+1}+ \frac{x^2+3-4}{\sqrt{x^2+3}+2}=x+1$
\Leftrightarrow$\frac{2x-2}{\sqrt{2x-1}+1}+ \frac{x^2-1}{\sqrt{x^2+3}+2}-(1-x)=0$
\Leftrightarrow$(x-1)(\frac{2}{\sqrt{2x-1}+1}+\frac{x+1}{\sqrt{x^2+3}+2}+1)=0$
* x-1=0\Leftrightarrow x=1(t/m)
*$\frac{2}{\sqrt{2x-1}+1}+\frac{x+1}{\sqrt{x^2+3}+2}+1=0$(2)
c/m vô nghiệm : (1) \Rightarrow x+1 > 0
\Rightarrow (2) > 0 \Rightarrow (1) vô nghiệm
Vậy pt có 1 nghiệm x= 1
 
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