Xét tam giác ABC có I là tâm đường tròn nội tiếp

\(\Rightarrow S_{ABC}=S_{AIB}+S_{BIC}+S_{CIA}=\frac{1}{2}AB.r+\frac{1}{2}BC.r+\frac{1}{2}CA.r\)

\(=\frac{1}{2}\left(AB+BC+CA\right).r=p.r\)

\(\Rightarrow r=\frac{S_{ABC}}{p}\)

\(\frac{S}{h_a}+\frac{S}{h_b}+\frac{S}{h_c}=\frac{1}{2}\left(a+b+c\right)=p=\frac{S}{r}\)

\(\Rightarrow\frac{1}{r}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\)

Học tốt!!!!!!!!!!!!!!!!

Lời giải :

\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}\)

Áp dụng AM-GM :

\(A\ge2\sqrt{2\sqrt{x}\cdot\frac{2}{\sqrt{x}}}=2\cdot2=4\)

Dấu "=" xảy ra \(\Leftrightarrow2\sqrt{x}=\frac{2}{\sqrt{x}}\Leftrightarrow x=1\)

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

Áp dụng ta có 

\(\frac{B}{2}>\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{24}-\sqrt{23}+\sqrt{25}-\sqrt{24}=5-1=4\)

=> \(B>8\)(ĐPCM)

Vậy B>8

\(a,7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}\)(  Với A>= 0,  B>=0)

\(=\left(7\sqrt{AB}-\sqrt{A}\right)+\left(7B-\sqrt{B}\right)\)

\(=7\sqrt{A}\left(\sqrt{B}-1\right)+7\sqrt{B}\left(\sqrt{B}-1\right)\)

\(=\left(\sqrt{B}-1\right)\left(7\sqrt{A}+7\sqrt{B}\right)\)

\(=7\left(\sqrt{B}-1\right)\left(\sqrt{A}+\sqrt{B}\right)\)

\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)Với a>= 0,  b>=0)

\(=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)

\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)

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

\(=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)

\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)

a, \(7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}=7\sqrt{B}\left(\sqrt{A}+\sqrt{B}\right)-\left(\sqrt{A}+\sqrt{B}\right)\)\(=\left(\sqrt{A}+\sqrt{B}\right)\left(7\sqrt{B}-1\right)\)

b, \(a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)

c,\(\sqrt{x^2-25y^2}-\sqrt{x-5y}=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)

\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)

Lời giải :

Xét dạng tổng quát sau : 

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

\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

Từ đó ta có hướng giải quyết bài toán :

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)

\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}\)

\(A=\sqrt{25}-\sqrt{1}\)

\(A=4\)

1) Ta có MP ^ 2 = 15^2 = 225 cm

Mà MN^2 + NP ^2 = 12^ 2 + 9^2 = 144 + 81 = 225

=> MP^2 = MN^2 + NP^2

=> Tam giác MNP cân tại N

kb vs mk nhé

Lời giải :

\(\sqrt{10}+\sqrt{5}+1>\sqrt{9}+\sqrt{4}+1=3+2+1=6\)

\(\sqrt{35}< \sqrt{36}=6\)

Từ đây ta có : \(\sqrt{10}+\sqrt{5}+1>6>\sqrt{35}\)

Vậy \(\sqrt{10}+\sqrt{5}+1>\sqrt{35}\)