+ Áp dụng BĐT Cô - si :
\(\sqrt{3x-9}=\frac{3.\sqrt{3x-9}}{3}=\frac{\frac{\sqrt{9.\left(3x-9\right)}}{2}}{3}=\frac{x}{2}\)

\(\sqrt{7-x}=\sqrt{1.\left(7-x\right)}\le\frac{1+7-x}{2}=\frac{8-x}{2}\)

Cộng theo vế ta được :

\(\sqrt{3x-9}+\sqrt{7-x}\le\frac{x+8-x}{2}=4\)

Dấu " = " xảy ra \(\Leftrightarrow x=6\)

Chúc bạn học tốt !!!

\(D=\dfrac{\sqrt{x}-x-7}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-9}{\sqrt{x}+1}=1-\sqrt{x}+1-\dfrac{9}{\sqrt{x}+1}\)

\(=3-\left[\left(\sqrt{x}+1\right)+\dfrac{9}{\sqrt{x}+1}\right]\)\(\le3-2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}\) ( BĐT AM-GM)

\(\Leftrightarrow D\le-3\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=\dfrac{9}{\sqrt{x}+1}\Leftrightarrow x=4\)

Vậy \(max_D=-3\)

ta có

\(\sqrt{\left(x-5\right).1}\le\frac{x-5+1}{2}=\frac{x-4}{2}\)

\(\sqrt{\left(7-x\right).1}\le\frac{7-x+1}{2}=\frac{-x+8}{2}\)

\(\Rightarrow P\ge\frac{x-4}{2}+\frac{8-x}{2}=2\)

Dấu = xảy ra <=> \(\hept{\begin{cases}x-5=1\\7-x=1\end{cases}\Leftrightarrow x=6}\)

vậy min P=2 khi x=6

- Đề sai nhiều vậy sửa lại đi bạn ;-;

e) Ta có: \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{2}{3}\right\}\)

ĐK: \(x\ge0\)

Mà: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\forall x\left(xđ\right)\\x\sqrt{x}\ge0\forall x\left(xđ\right)\end{matrix}\right.\)

Nên: \(M=\dfrac{\sqrt{x}+1}{x\sqrt{x}+7}\le\dfrac{1}{7}\forall x\left(xđ\right)\)

Dấu "=" xảy ra:

\(\dfrac{\sqrt{x}+1}{x\sqrt{x}+7}=\dfrac{1}{7}\)

\(\Leftrightarrow7\sqrt{x}+7=x\sqrt{x}+7\)

\(\Leftrightarrow x\sqrt{x}-7\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\left(tm\right)\)

Vậy: \(M_{max}=\dfrac{1}{7}\) khi \(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)