[Toán 10]phương trình vô tỷ!

L

lequan19942011

bài 2
đặt[TEX] \sqrt[2]{3x^2-2x-3} =a [/TEX]
đặt x+6=b
[TEX]\Rightarrow a^2+5b=ab+25[/TEX]
[TEX]\Leftrightarrow (a-5)(a+5-b)=0[/TEX]
[TEX]\Leftrightarrow[/TEX] hoặc a=5 hoặc a=b-5
th1 [TEX]a=5 \Leftrightarrow \sqrt[2]{3x^2-2x-3}=5[/TEX]
[TEX]\Leftrightarrow 3x^2-2x-3=25 \Leftrightarrow 3x^2-2x-28=0[/TEX]

th2 [TEX]a=5+b \Leftrightarrow 3x^2-2x-3=x^2+2x+1 \Leftrightarrow 2x^2-4x-4=0[/TEX]
:D
 
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0

01263812493

Giải phương trình:
a, [TEX]x^2+x+2=(3x-2).\sqrt{x+1}(')[/TEX]
c,[TEX]\sqrt{x^2+x+2}(3x+1)=3x^2+3x+2('')[/TEX]

[TEX]\blue a) dk: \ x > \frac{2}{3}[/TEX]
[TEX]\blue (') \leftrightarrow (x^2+x+2)^2=(9x^2-12x+4)(x+1)[/TEX]
[TEX]\blue \leftrightarrow x(x-3)(x^2-4x-4)=0 [/TEX]

[TEX]\blue c) dk: \ x > \frac{-1}{3}[/TEX]
[TEX]\blue ('') \leftrightarrow \sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2[/TEX]
[TEX]\blue \leftrightarrow \frac{(x-1)(x+2)}{\sqrt{x^2+x+2}+2}=\frac{3x(x-1)}{3x+1}[/TEX]

[TEX]\blue \rightarrow \left[x=1 (n)\\ \frac{x+2}{\sqrt{x^2+x+2}+2}=\frac{3x}{3x+1}(+)[/TEX]
Giải (+) ta có:
[TEX]\blue (+) \leftrightarrow 3x^2+x+2=3x\sqrt{x^2+x+2}[/TEX]
Đặt: [TEX]\blue a=\sqrt{x^2+x+2} >0[/TEX]
[TEX]\blue 2x^2-3ax+a^2=0[/TEX]
[TEX]\blue \Delta =a^2 \Rightarrow ....[/TEX]
 
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