giỏi ha

hay ghê nhờ

\(Ta\)\(có\)\(:\)

\(tana\)\(=\frac{HM}{AH}\)

\(\Rightarrow2\)\(tana\)\(=\frac{2HM}{AH}\)\(=\frac{CH-BH}{AH}\)\(=\frac{CH}{AH}\)\(-\frac{BH}{AH}\)

\(\Rightarrow cot\)\(C\)\(-\)\(cot\)\(B\)

\(\Rightarrow\)\(tana\)\(=\frac{cotC-cotB}{2}\)

\(\frac{\sqrt{13,5}}{\sqrt{4,5}}=\sqrt{\frac{13,5}{4,5}}=\sqrt{3}\)

Ta có :  \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=6\)

\(\Rightarrow\left(a+b+c\right)\cdot\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)=6.\left(a+b+c\right)\)

\(\Leftrightarrow\frac{\left(a+b+c\right)\cdot\left(a+b\right)}{c}+\frac{\left(a+b+c\right)\cdot\left(b+c\right)}{a}+\frac{\left(a+b+c\right)\cdot\left(c+a\right)}{b}=24\) ( Do \(a+b+c=4\) )

\(\Leftrightarrow\frac{\left(a+b\right)^2+c.\left(a+b\right)}{c}+\frac{\left(b+c\right)^2+a.\left(b+c\right)}{a}+\frac{\left(c+a\right)^2+b.\left(c+a\right)}{b}=24\)

\(\Leftrightarrow\left[\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\right]+2\left(a+b+c\right)=24\)

\(\Leftrightarrow\left[\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\right]+2.4=24\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}=16\) ( đpcm )

Bài làm:

Xét: \(\sqrt{2}< \sqrt{11+6\sqrt{2}}\)

=> \(\sqrt{2}-\sqrt{11+6\sqrt{2}}< 0\) (1)

và \(5>\sqrt{5}\) => \(5-\sqrt{5}>0\)

<=> \(2\sqrt{5-\sqrt{5}}>0\) => \(\sqrt{6+2\sqrt{5-\sqrt{5}}}>0\) (2)

Từ (1) và (2)

=> \(\frac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5-\sqrt{5}}}}< 0\)

Mà biểu thức trong căn phải có giá trị không âm

=> Mâu thuẫn

=> Căn thức không có giá trị

\(\sqrt{85}.\sqrt{125}.\sqrt{68}=\sqrt{85.125.68}=\sqrt{5.17.5.25.17.4}\)

\(=\sqrt{5^2.25.17^2.4}=\sqrt{5^2}.\sqrt{25}.\sqrt{17^2}.\sqrt{4}=5.5.17.2=850\)