a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)

\(\Leftrightarrow\sqrt{x}+1\ge1\)

\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)

\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)

\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi x=0

Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0

Ta có : \(\left\{{}\begin{matrix}x\ge1\\y\ge2\\z\ge3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{y-2}\ge0\\\sqrt{z-3}\ge0\end{matrix}\right.\Rightarrow\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}\ge0\)

Đặt \(\sqrt{x-1}=a;\sqrt{y-2}=b;\sqrt{z-3}=c\)

\(\Rightarrow A=\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\)

\(\sum\frac{a}{a^2+1}=\sum\left(a-\frac{a^3}{a^2+1}\right)\ge\sum\left(a-\frac{a}{2}\right)=\frac{a+b+c}{2}\)

\(\Rightarrow A\ge\frac{\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}}{2}=0\)

Vậy \(MIN_A=0\) khi \(x=1;y=2;z=3\)

\(A=\frac{1.\sqrt{x-1}}{x}+\frac{1}{\sqrt{2}}.\frac{\sqrt{2}.\sqrt{y-2}}{y}+\frac{1}{\sqrt{3}}.\frac{\sqrt{3}.\sqrt{z-3}}{z}\)

\(A\ge\frac{1+x-1}{2x}+\frac{1}{\sqrt{2}}\left(\frac{2+y-2}{2y}\right)+\frac{1}{\sqrt{3}}\left(\frac{3+z-3}{2z}\right)=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\)

\(\Rightarrow A_{min}=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

A=
x
1.
x−1
​

​
+
2
​

1
​
.
y
2
​
.
y−2
​

​
+
3
​

1
​
.
z
3
​
.
z−3
​

​


A\ge\frac{1+x-1}{2x}+\frac{1}{\sqrt{2}}\left(\frac{2+y-2}{2y}\right)+\frac{1}{\sqrt{3}}\left(\frac{3+z-3}{2z}\right)=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}A≥
2x
1+x−1
​
+
2
​

1
​
(
2y
2+y−2
​
)+
3
​

1
​
(
2z
3+z−3
​
)=
12
6+3
2
​
+2
3
​

​


\Rightarrow A_{min}=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}⇒A
min
​
=
12
6+3
2
​
+2
3
​

​
khi \left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.
⎩
⎨
⎧
​

x−1
​
=1
y−2
​
=
2
​

z−3
​
=
3
​

​
\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.⇒
⎩
⎨
⎧
​

x=2
y=4
z=6
​

1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)

Đạt được khi x = 9

2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)

\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)

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

Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)

Không có GTLN nhé

3/ Điều kiện xác định bạn tự làm nhé

\(\frac{16}{\sqrt{x}+3}=\frac{-8\sqrt{x}+5}{3\sqrt{x}+1}\)

\(\Leftrightarrow8x+67\sqrt{x}+1=0\)

Tới đây thì bạn xem như phương trình bậc 2 là giải tiếp được. Nhớ đối chiếu điều kiện để loại nghiệm