can gap tiep

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khongcobg

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pe_candy_sweet1995

1) [TEX]\frac{x^2}{\sqrt{3x-2}[/TEX] - [TEX]\sqrt{3x-2}=1-x[/TEX]

bai nay giai nhu the nay nhe

[TEX]\frac {x^2}{\sqrt{3x-2}[/tex]= 1-x
= [TEX]\frac{x^2-3x +2 }{\sqrt{3x-2}[/tex]=-(x-1)
=[tex] x^2 -3x +2 =-(x-1)[tex]\sqrt{3x-2}[/tex]
=(x-1)(x-2) =-(x-1)[tex]\sqrt{3x-2}[/tex]
=(x-1)(x-2) +(x-1)[tex]\sqrt{3x-2}=0[/tex]
=(x-1)(x-2+[tex]\sqrt{3x-2})=0 =>[TEX]\left[\begin{x=1}\\{x-2 + [TEX]\sqrt{3x-2}= 0}[/TEX]
=> 2-x= [TEX]\sqrt{3x-2} [/TEX]
=>[tex] (2-x)^2 [/tex]=3x-2
[tex] x^2 -4x +4 = 3x-2 [tex] x^2 -7x +6 =0 [TEX]\left[\begin{x1=1}\\{x2 = 6} [/TEX]
 
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0915549009

4) [TEX]\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1[/TEX]
Mình làm nốt bài 4 :):):):)
Ta có: [TEX]\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1 \Rightarrow 5 - 2\sqrt{(3-x+x^2)(2+x-x^2)} = 1 \Leftrightarrow - 2\sqrt{(3-x+x^2)(2+x-x^2)} = -4 \Leftrightarrow \sqrt{(3-x+x^2)(2+x-x^2)} = 2 \Leftrightarrow (3-x+x^2)(2+x-x^2) = 4 \Leftrightarrow - x^4 + 2x^3 - 2x^2 + x + 2 = 0[/TEX]
Sau đó bạn phân tích đa thức thành nhân tử là giải đc thui.
 
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