a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

Em đã thử liên hợp nhưng cái ngoặc to xấu xí quá:(

pt <=> \(2x^2-20x+54-2\sqrt{x-4}-2\sqrt{6-x}=0\)

<=> \(\left(2x^2-20x+50\right)+\left(x-4-2\sqrt{x-4}+1\right)+\left(6-x-2\sqrt{6-x}+1\right)=0\)

<=> \(2\left(x-5\right)^2+\left(\sqrt{x-4}-1\right)^2+\left(\sqrt{6-x}-1\right)^2=0\)

<=> x = 5

mik ko biết vì mới chỉ học lớp 6

ĐKXĐ: \(x\ge\frac{1}{2}\)

Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)

Nhân liên hợp ta được:

\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)

mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)

=> x - 2 = 0 => x = 2

                                                   Vậy x = 2

rảnh  quá 

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

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

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

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

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

\(\Leftrightarrow x=2\)

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

\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)

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

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

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

ta có đề bài <=> 

\(\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\)

<=> \(\left|x-3\right|+\left|x+5\right|=8\)

<=>\(\left|3-x\right|+\left|x+5\right|=8\)

Áp dụng tính chât dấu giá trị tuyệt đối ta có 

\(\left|3-x\right|+\left|x+5\right|>=\left|3-x+x+5\right|=8\)

dấu = xảy ra <=> \(\left(3-x\right)\left(x+5\right)>=0\)

đến đây bạn tự giaỉ dấu = nhé

Lời giải:

Với mọi $x$ thuộc ĐKXĐ, ta luôn có:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}\geq 0\\ \sqrt{x^2+3x+1}\geq 0\end{matrix}\right.\)

Do đó, để \(\sqrt{3x+x^2+\frac{9}{4}}+\sqrt{x^2+3x+1}=0\) thì:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}= 0\\ \sqrt{x^2+3x+1}=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=\frac{-3}{2}\\ x=\frac{3\pm \sqrt{5}}{2}\end{matrix}\right.\) (vô lý)

Do đó pt vô nghiệm.