ĐK: \(x\ge-2\)

\(pt\Leftrightarrow\frac{x+5-\left(x+2\right)}{\sqrt{x+5}+\sqrt{x+2}}.\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)=3\)

\(\Leftrightarrow3.\frac{1+\sqrt{x+2}.\sqrt{x+5}}{\sqrt{x+2}+\sqrt{x+5}}=3\)

\(\Leftrightarrow1+\sqrt{x+2}\sqrt{x+5}=\sqrt{x+2}+\sqrt{x+5}\)

\(\Leftrightarrow\left(\sqrt{x+2}-1\right)\left(\sqrt{x+5}-1\right)=0\)

\(\Leftrightarrow\sqrt{x+2}=1\text{ hoặc }\sqrt{x+5}=1\)

\(\Leftrightarrow x=-1\text{ (nhận) hoặc }x=-4\text{ (loại)}\)

Vậy tập nghiệm của pt là: \(S=\left\{1\right\}\)

\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)

\(\Leftrightarrow\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)

\(\Rightarrow\left(x-2\right)^2=x^2-4\)

\(\Leftrightarrow x^2-4x+4-x^2+4=0\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow x=2\)

Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{x-2}=b;\) ta có:

\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)

\(\Leftrightarrow2a^2+ab-2ab-b^2=0\)

⇔ \(\left(2a+b\right)\left(a-b\right)=0\)

⇔ \(\left[{}\begin{matrix}2\sqrt[3]{x+2}=-\sqrt[3]{x-2}\\\sqrt[3]{x-2}=\sqrt[3]{x+2}\end{matrix}\right.\)⇔ \(x=-\frac{14}{9}\)

X = 0

100% đúng kq

x=0

nhanh nhất có thể

b/ Đặt \(\sqrt{x^2+1}=a\ge0\)

\(\Rightarrow a^2+3x=\left(x+3\right)a\)

\(\Leftrightarrow\left(3-a\right)\left(x-a\right)=0\)

a/ Dựa vô TXĐ thì thấy \(x< 2\)

\(\Rightarrow\sqrt{x^2+6}+2\sqrt{x^2-1}-x>\sqrt{6}-2>0\)

Vậy vô nghiệm

Lời giải:

ĐK: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow \sqrt{2x+3}=2x^2-5x$

$\Leftrightarrow \sqrt{2x+3}-3=2x^2-5x-3$

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

$\Leftrightarrow (x-3)\left[\frac{2}{\sqrt{2x+3}+3}-(2x+1)\right]=0$

Xảy ra 2 TH:

TH1: $x-3=0\Rightarrow x=3$ (thỏa mãn)

TH2: $\frac{2}{\sqrt{2x+3}+3}=2x+1$

Đặt $\sqrt{2x+3}=t(t\geq 0)$ thì pt trở thành: \frac{2}{t+3}=t^2-2$

$\Leftrightarrow 2=(t^2-2)(t+3)\Leftrightarrow t^3+3t^2-2t-8=0$

$\Leftrightarrow (t+2)(t^2+t-4)=0$

Do $t\geq 0$ nên $t=\frac{-1+\sqrt{17}}{2}$

$\Leftrightarrow \sqrt{2x+3}=\frac{-1+\sqrt{17}}{2}\Leftrightarrow x=\frac{3-\sqrt{17}}{4}$ (thỏa mãn)

Vậy........