\(P=\frac{\left(x+7\right)}{\sqrt{x}+3}=\)\(\frac{\left(x+3\sqrt{x}\right)-3\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}\)\(=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}\)

\(=\left(\sqrt{x}+3\right)+\frac{16}{\sqrt{x}+3}-6\)\(\ge8-6=2\)(AM-GM)

''='' <=> x = 1

P\(=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\)

ap dung cosi cho 2 so duong \(\left(\sqrt{x}+3\right)va\frac{16}{\sqrt{x}+3}taduoc\)

\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\frac{16}{\sqrt{x}+3}}\)\(\ge2\sqrt{16}=8\)

\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\ge8-3=5\)

dau = xay ra <=> \(\left(\sqrt{x}+3\right)=\frac{16}{\sqrt{x}+3}\)

<=>  x=1

Áp dụng BĐT \(\sqrt{a^2+b^2}\ge\frac{\sqrt{2}}{2}\left(a+b\right)\) (bạn tự chứng minh)

Ta có \(P=\frac{\sqrt{x^2+y^2}}{z}+\frac{\sqrt{y^2+z^2}}{x}+\frac{\sqrt{z^2+x^2}}{y}\ge\frac{\sqrt{2}}{2}\left(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\right)\)

\(=\frac{\sqrt{2}}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\right]\ge\frac{\sqrt{2}}{2}\left(2+2+2\right)=3\sqrt{2}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x=y=z\\x,y,z>0\end{cases}}\)

Vậy min P = \(3\sqrt{2}\) khi x = y = z

a) \(=x+4+\frac{25}{x+4}-4\). x>-4 => x+4>0. => 25/x+4 >0

áp dụng bđt cosi  cho 2 số dương ta có: \(x+4+\frac{25}{x+4}\ge2\sqrt{\left(x+4\right).\frac{25}{x+4}}=2\sqrt{25}=10\Rightarrow x+4+\frac{25}{x+4}-4\ge10-4=6\)

=> GTNN=6  <=> x=1

b) ĐK: x>=0, x khác 9

 \(A=\frac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)

tương tự ở trên để c/m 2 số dương rồi áp dụng bđt cosi \(A\ge2\sqrt{5}-6=4\)=> Min =4 <=> x=4

nếu vẫn k làm đc thì liên hệ mình mình giải nốt cho nha.

c) gọi là B đi. B=|x-3|+|x-5|

ta sẽ có bảng xét dấu:

Nếu \(x\le3\) <=> B=-x+3-x+5=-2x+8

x=<3 <=>-2x>-6 <=> -2x+8>2 <=> B>=2

Nếu 3<x<5 => B=x+3-x+5=0x+15=15=> B=15

Nếu x>=5=> B=x+3+x+5=2x+8 

x>=5 <=> 2x>10 <=>2x+8>=18 <=> B>=18

=> Min B=2 <=> x=3

nhớ LI KE

\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

\(=x-\sqrt{x}\)

\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)

Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)

\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)

c, Đề thiếu không bạn?

a)\(x+\frac{25}{x+4}=x+4+\frac{25}{x+4}-4\ge2\sqrt{\left(x+4\right).\frac{25}{x+4}}-4\)(Cô-si)

                                                             \(=2.5-4=6\)

Vậy: GTNN là 6 \(\Leftrightarrow x+4=\frac{25}{x+4}\Leftrightarrow x=1\)(do x >-4)

b)\(A=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)

        \(=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}-6=2.5-6=4\)

Vậy: A min = 4 <=> x = 4

c) Áp dụng bdt: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Dấu "=" xảy ra \(\Leftrightarrow ab\ge0\)

Ta có: \(\left|x-3\right|+\left|x-5\right|=\left|3-x\right|+\left|x-5\right|\ge\left|3-x+x-5\right|=2\)

\("="\Leftrightarrow\left(3-x\right)\left(x-5\right)\ge0\Leftrightarrow3\le x\le5\)