giải phương trình $x^{3}$+6 $x^{2}$+5$x^{}$-3-(2$x^{}$+5) $\sqrt[2]{2x+3}$=0

Đáp án: $\sqrt{2}$

Giải thích các bước giải:

ĐKXĐ: $x\ge-\dfrac32$

Ta có:

$x^3+6x^2+5x-3-(2x+5)\sqrt{2x+3}=0$

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

$\to x^3+6x^2+12x+8-7x-11=(2x+5)\sqrt{2x+3}$

$\to (x+2)^3-7x-11=(2x+3)\sqrt{2x+3}+2\sqrt{2x+3}$

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

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

Đặt $x+2=a, \sqrt{2x+3}+1=b$

Vì$ x\ge-\dfrac32\to a\ge \dfrac12, b\ge 1$

$\to a^3-a=b^3-b$

$\to (a^3-b^3)-(a-b)=0$

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

$\to (a-b)(a^2+ab+b^2-1)=0$

Ta có $a\ge \dfrac12,b\ge 1\to a^2+ab+b^2-1\ge \dfrac34>0$

$\to a-b=0$

$\to a=b$

$\to x+2=\sqrt{2x+3}+1$

$\to x+1=\sqrt{2x+3}$

$\to 2x+2=2\sqrt{2x+3}$

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

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

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

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

$\to \sqrt{2x+3}-1=\pm\sqrt{2}$

$\to \sqrt{2x+3}=1\pm\sqrt{2}$

Vì $\sqrt{2x+3}\ge 0$

$\to \sqrt{2x+3}=1+\sqrt{2}$

$\to x=\sqrt{2}$