Giả pt: a) 2x ² – 5x = ( x -3). $\sqrt[2]{3x-2}$ + (x – 2) . $\sqrt[2]{4x-3}$ b) ( x ² + x). $\sqrt[2]{2x+3}$ = x ³ + 3x ² + x – 2 (Giúp mình 1 t

0 bình luận về “Giả pt: a) 2x ² – 5x = ( x -3). $\sqrt[2]{3x-2}$ + (x – 2) . $\sqrt[2]{4x-3}$ b) ( x ² + x). $\sqrt[2]{2x+3}$ = x ³ + 3x ² + x – 2 (Giúp mình 1 t”

1. Giải thích các bước giải:

   a.Ta có :

   $2x^2-5x=(x-3)\sqrt{3x-2}+(x-2)\sqrt{4x-3}$

   $\to 2x^2-5x-x(x-3)-x(x-2)=(x-3)(\sqrt{3x-2}-x)+(x-2)(\sqrt{4x-3}-x)$

   $\to(x-3).\dfrac{3x-2-x^2}{\sqrt{3x-2}+x}+(x-2).\dfrac{4x-3-x^2}{\sqrt{4x-3}+x}=0$

   $\to(x-3).\dfrac{-(x-1)(x-2)}{\sqrt{3x-2}+x}+(x-2).\dfrac{-(x-1)(x-3)}{\sqrt{4x-3}+x}=0$

   $\to\dfrac{(x-1)(x-2)(x-3)}{\sqrt{3x-2}+x}+\dfrac{(x-1)(x-2)(x-3)}{\sqrt{4x-3}+x}=0$

   $\to (x-1)(x-2)(x-3)(\dfrac{1}{\sqrt{3x-2}+x}+\dfrac{1}{\sqrt{4x-3}+x})=0$

   $\to (x-1)(x-2)(x-3)=0$ vì $x\ge \dfrac34$

   $\to x\in\{1,2,3\}$

   b.Ta có :

   $(x^2+x)\sqrt{2x+3}=x^3+3x^2+x-2$

   Đặt $\sqrt{2x+3}=a,x+1=b, a\ge 0$

   $\to (x^2+x)\sqrt{2x+3}=x^3+3x^2+x-2$

   $\to x(x+1)\sqrt{2x+3}=(x+1)^3-(2x+3)$

   $\to (x^2+x)\sqrt{2x+3}=(x+1)^3-(2x+3)$

   $\to ((x+1)^2-(x+1))\sqrt{2x+3}=(x+1)^3-(2x+3)$

   $\to (b^2-b)a=b^3-a^2$

   $\to ab^2-ab-b^3+a^2=0$

   $\to b^2(a-b)+a(a-b)=0$

   $\to (b^2+a)(a-b)=0$

   $\to a-b=0\to a=b\to \sqrt{2x+3}=x+1\to x=\pm\sqrt2$

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