giải phương trình sau: 2(2x-1) – 3. √(5x-6)= √(3x-8)

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1. 2(2x – 1) – 3√(5x – 6) = √(3x – 8)
   x ∈ [ 8  ; ∞ ]
           3
   2(2x – 1) – 3√(5x – 6) = ( 3√( 5x – 6 ) – 4x + 2 )
   ( 3√( 5x – 6 ) – 4x + 2 ) = √( 3x – 8 )
   – 3 √( 5x – 6 ) – √( 3x – 8 ) + 4x – 2 = 0
   <=>  x = 3

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2. Đáp án: $x=3$

   Giải thích các bước giải:

   ĐKXĐ: $x\ge \dfrac83$

   Ta có :
   $2(2x-1)-3\sqrt{5x-6}=\sqrt{3x-8}$

   $\to 2(2x-1)=3\sqrt{5x-6}+\sqrt{3x-8}$

   $\to 2(2x-1)-10=(3\sqrt{5x-6}-9)+(\sqrt{3x-8}-1)$

   $\to 2(2x-1-5)=3(\sqrt{5x-6}-3)+(\sqrt{3x-8}-1)$

   $\to 2(2x-6)=3\cdot\dfrac{5x-6-3^2}{\sqrt{5x-6}+3}+\dfrac{3x-8-1}{\sqrt{3x-8}+1}$

   $\to 4(x-3)=3\cdot\dfrac{5(x-3)}{\sqrt{5x-6}+3}+\dfrac{3(x-3)}{\sqrt{3x-8}+1}$

   $\to x-3=0\to x=3$

   Hoặc $\to 4=3\cdot\dfrac{5}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}$

   $\to 4=\dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}(1)$

   $+) x>3$

   $\to \dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}<\dfrac{15}{\sqrt{5\cdot3-6}+3}+\dfrac{3}{\sqrt{3\cdot 3-8}+1}=4$

   $\to (1)$ vô nghiệm

   $+) x<3\to \dfrac83\le x<3$

   $\to \dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}>\dfrac{15}{\sqrt{5\cdot3-6}+3}+\dfrac{3}{\sqrt{3\cdot 3-8}+1}=4$

   $\to (1)$ vô nghiệm

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