giải phương trình sau

L

lethiquynhhien

ĐK: [TEX]x\geq \frac{3}{2}[/TEX]
[TEX]PT \Leftrightarrow \frac{1}{\sqrt[]{3x}}+\frac{1}{\sqrt[]{6x-9}}=\frac{1}{\sqrt[]{4x-3}}+\frac{1}{\sqrt[]{5x-6}}[/TEX]
[TEX]\Leftrightarrow \frac{(\sqrt[]{3x}+\sqrt[]{6x-9})^2}{3x(6x-9)}=\frac{(\sqrt[]{5x+6}+\sqrt[]{4x-3})^2}{(4x-3)(5x-6)}[/TEX]
[TEX]\Leftrightarrow \frac{9(x-9)}{9x(2x-3)} +\frac{2}{\sqrt[]{9x(2x-3)}}=\frac{9(x-9)}{(4x-3)(5x-6)}+\frac{2}{\sqrt[]{(4x-3)(5x-6)}}[/TEX]
đặt [TEX]\sqrt[]{9x(2x-3)}=a , \sqrt[]{(4x-3)(5x-6)}=b[/TEX], ta có:
[TEX]9(x-1)(\frac{1}{a^2}-\frac{1}{b^2})=2(\frac{1}{b} - \frac{1}{a})\Rightarrow a=b [/TEX]hoặc 9(x-1)(1/a+1/b)=-2(loại do a,b>0 và [TEX]x \geq 3/2[/TEX])
 
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B

botvit

[TEX]PT \Leftrightarrow \frac{1}{\sqrt[]{3x}}+\frac{1}{\sqrt[]{6x-9}}=\frac{1}{\sqrt[]{4x-3}}+\frac{1}{\sqrt[]{5x-6}}[/TEX]
[TEX]\Leftrightarrow \frac{(\sqrt[]{3x}+\sqrt[]{6x-9})^2}{3x(6x-9)}=\frac{(\sqrt[]{5x+6}+\sqrt[]{4x-3})^2}{(4x-3)(5x-6)}[/TEX]
[TEX]\Leftrightarrow \frac{9(x-9)}{9x(2x-3)} +\frac{2}{\sqrt[]{9x(2x-3)}}=\frac{9(x-9)}{(4x-3)(5x-6)}+\frac{2}{\sqrt[]{(4x-3)(5x-6)}}[/TEX]
đặt [TEX]\sqrt[]{9x(2x-3)}=a , \sqrt[]{(4x-3)(5x-6)}=b[/TEX], ta có:
[TEX]9(x-1)(\frac{1}{a^2}-\frac{1}{b^2}=2(\frac{1}{b}-\frac{1}{a} \Rightarrow a=b [/TEX]hoặc 9(x-1)(1/a+1/b)=-2(loại do a,b>0 và x[TEX]\geq[/TEX] 3/2
phải = 2 chứ sao -2?
(1/a-1/b)(9(x-1)(1/a+1/b)=2
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