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

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

''='' <=> x = 1

\(Q=\sqrt{x}+\dfrac{25}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{25}{\sqrt{x}}}=10\)

Dấu = xảy ra khi x=25

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:  \(x>0;x\ne9\)

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

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

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

\(=\frac{1}{\sqrt{x}+3}\)

b)  \(A=\frac{1}{5}\) \(\Rightarrow\)\(\frac{1}{\sqrt{x}+3}=\frac{1}{5}\)

\(\Rightarrow\)\(\sqrt{x}+3=5\)

\(\Leftrightarrow\)\(\sqrt{x}=2\)

\(\Leftrightarrow\)\(x=4\)(t/m ĐKXĐ)

Vậy...

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

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