1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

ĐKXĐ : \(1\le x\le3\)

Ta có \(\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}\ge x^3+10\)

<=> \(-2\sqrt{x-1}-2\sqrt{3-x}-8x\sqrt{2x}\le-2x^3-20\)

<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+2x^3-8x\sqrt{2x}+16\le0\)(1)

Đặt \(\sqrt{2x}=y\) => \(x=\dfrac{y^2}{2}\)

Khi đó \(2x^3-8x\sqrt{2x}+16=\dfrac{y^6}{4}-4y^3+16=\left(\dfrac{y^3-8}{2}\right)^2\)

Khi đó (1) <=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\le0\)(1)

mà \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\ge0\forall x;y\)(2) 

Từ (2)(1) => \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2=0\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{3-x}-1=0\\\dfrac{y^3-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\3-x=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\\\sqrt{2x}=2\end{matrix}\right.\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm bất phương trình

mình nghĩ sửa đề bài là  \(\frac{\sqrt{x^2-x+6}+7\sqrt{x}-\sqrt{6\left(x^2+5x-2\right)}}{x+3-\sqrt{2\left(x^2+10\right)}}\le0\) 

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)