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 trong 2 câu cũng được nha..mong là giúp mình cả.:D mình sẽ vote 5 sao..) cảm ơn trước.!

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 $\sqrt{2x+3}=a,x+1=b, a\ge 0$

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

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

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

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

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

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

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

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

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