\(P=\frac{5a+5b+2c}{\sqrt{12\left(a^2+ab+bc+ca\right)}+\sqrt{12\left(b^2+ab+bc+ca\right)}+\sqrt{c^2+ab+bc+ca}}\)

\(=\frac{5a+5b+2c}{\sqrt{12\left(a+b\right)\left(a+c\right)}+\sqrt{12\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(=\frac{5a+5b+2c}{\sqrt{\left(6a+6b\right)\left(2a+2c\right)}+\sqrt{\left(6a+6b\right)\left(2b+2c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{6a+6b+2a+2c+6a+6b+2b+2c+a+c+b+c}\)

\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{3\left(5a+5b+2c\right)}=\frac{2}{3}\)

\(P_{min}=\frac{2}{3}\) khi \(\left\{{}\begin{matrix}a=b=1\\c=5\end{matrix}\right.\)

thay 28 vào pt nhân tử rồi cối dưới mẫu

thay 28 vao pt nhan tu roi am-gm cho cai do luon

Ps: tim Min

Bài 1:

a)    \(=5.|2a|-5a^2\)

b)    \(=7\left(a-1\right)+5a=12a-7\)

c)    \(|a-2|-5\sqrt{a+2}\)

Bài 2:

a)    \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)

b)    \(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)

\(=2\sqrt{2}\)

c)    \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)

\(=-2\sqrt{5}\)

a) \(5\sqrt{4a^2}-5a^2\)

\(=5.|2a|-5a^2\)

b) \(7\sqrt{\left(a-1\right)^2}+5a\)

\(=7\left(a-1\right)+5a\)

\(=12a-7\)

c) \(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\)

\(=|a-2|-5\sqrt{a+2}\)

bài 2:

a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)

\(=3-\sqrt{2}+5-\sqrt{2}\)

\(=8-2\sqrt{2}\)

b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)

\(=2\sqrt{2}\)

c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)

\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)

\(=-2\sqrt{5}\)

Bài 1 : 

a, \(5\sqrt{4a^2}-5a^2=5\sqrt{2^2a^2}-5a^2=5.2a-5a^2=10a-5a\)

b, \(7\sqrt{\left(a-1\right)^2}+5a=7a-7+5a=12a-7\)

c.\(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}=2-a-5\sqrt{a+2}\)

Mẫu thức như vầy thì tìm max còn được chứ tìm min sao nổi bạn?

\(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+30\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)

\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)

\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{40}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2007\)

\(\Leftrightarrow\frac{5}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le2007\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{\frac{6021}{5}}\)

Ta có:

\(5a^2+2ab+2b^2=4a^2+2ab+b^2+a^2+b^2\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)

\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}\)

\(\Rightarrow P\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{3}\sqrt{\frac{6021}{5}}\)

Dấu "=" xảy ra khi \(a=b=c=3\sqrt{\frac{5}{6021}}\)

\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)

Ta có:

\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)

C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)

C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)

\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)

sao dấu= thứ 2 lại ra như vậy

sai ở bc 2 rồi