Bài 11 :

Gọi \(x,y,z\left(m\right)\) lần lượt là cạnh \(AB,AC,BC\) của tam giác ABC

\(\left(0< x,y,z< 12\right)\)

Theo đề bài, ta có hệ pt :

\(\left\{{}\begin{matrix}x+y+z=12\left(1\right)\\x^2+y^2+z^2=50\left(2\right)\end{matrix}\right.\)

Theo d/l Pytago : \(x^2+y^2=z^2\) \(\left(3\right)\)

\(\left(2\right),\left(3\right)\Rightarrow z^2+z^2=50\Rightarrow2z^2=50\Rightarrow z=5\left(tmdk\right)\)

Thay \(z=5\) vào hệ pt \(\left(1\right),\left(2\right)\) ta có :

\(\left\{{}\begin{matrix}x+y+5=12\\x^2+y^2+5^2=50\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\x^2+y^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7-y\\\left(7-y\right)^2+y^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7-y\\49-14y+y^2+y^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7-y\\2y^2-14y+24=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7-y\\y^2-7y+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7-y\left(4\right)\\\left\{{}\begin{matrix}y=4\left(tmdk\right)\\y=3\left(tmdk\right)\end{matrix}\right.\end{matrix}\right.\)

Thay \(y=4\) vào \(\left(4\right)\) ta có \(x=7-4=3\left(tmdk\right)\)

Thay \(y=3\) vào \(\left(4\right)\) ta có \(x=7-3\left(tmdk\right)\)

Vậy độ dài cạnh AB,AC,BC lần lượt là \(3cm,4cm,5cm\) hoặc \(4cm,3cm,5cm\)

Sửa : Phần KL là m chứ không phải cm nha.

a: \(\sqrt{12+2\sqrt{35}}=\sqrt{7}+\sqrt{5}\)

b: \(\sqrt{16+6\sqrt{7}}=3+\sqrt{7}\)

d: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

e: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ge0\\x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có : \(B=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{2\sqrt{x}}\)

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

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

=> \(B=\frac{1}{\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{2}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{3}{2\sqrt{x}}\)

Vậy ....

ủa mà sao lại thank guy dạ

B

AD sao = AC đc nó có đi qua tâm đâu bạn

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)

\(\Leftrightarrow\dfrac{1105-1100}{x+5}=2\)

\(\Leftrightarrow\dfrac{5}{x-5}=2\)

\(\Leftrightarrow5=2\left(x-5\right)\)

\(\Leftrightarrow5=2x-10\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\dfrac{15}{2}=7,5\)

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(ĐK:x\ne0;x\ne-5\right)\\ \Leftrightarrow\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\\ \Leftrightarrow2x^2+10x-5500=0\\ \Leftrightarrow2x^2-100x+110x-5500=0\\ \Leftrightarrow2x.\left(x-50\right)+110.\left(x-50\right)=0\\ \Leftrightarrow\left(2x+110\right).\left(x-50\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+110=0\\x-50=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-55\left(TM\right)\\x=50\left(TM\right)\end{matrix}\right.\)

Vậy: S={-55;50}

\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

\(\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(3-a\right)^2}\)

\(=\left|a+3\right|+\left|3-a\right|=3+a+3-a=6\)

Bài 2: 

a: \(\dfrac{\sqrt{5}}{\sqrt{3}}=\dfrac{\sqrt{15}}{3}\)

b: \(\dfrac{13}{5-\sqrt{3}}=\dfrac{65+13\sqrt{3}}{22}\)

c: \(\dfrac{2\sqrt{2}}{\sqrt{7}+\sqrt{3}}=\dfrac{2\sqrt{14}-2\sqrt{6}}{4}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)

d: \(\dfrac{5-\sqrt{3}}{\sqrt{6}-2\sqrt{2}}=\dfrac{-3\sqrt{6}-7\sqrt{2}}{2}\)

Bài 3:

a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)

\(=\sqrt{3}+1-\sqrt{3}+1\)

=2

b: \(\dfrac{1}{3-\sqrt{3}}+\dfrac{1}{3+\sqrt{2}}\)

\(=\dfrac{3+\sqrt{3}}{6}+\dfrac{3-\sqrt{2}}{7}\)

\(=\dfrac{21+7\sqrt{3}+18-6\sqrt{2}}{42}\)

\(=\dfrac{39+7\sqrt{3}-6\sqrt{2}}{42}\)

Xét (O) có

MA là tiếp tuyến có A là tiếp điểm

MB là tiếp tuyến có B là tiếp điểm

Do đó: MA=MB

Xét (O) có 

NA là tiếp tuyến có A là tiếp điểm

NC là tiếp tuyến có C là tiếp điểm

Do đó: NA=NC

Ta có: MN=NA+MA

nên MN=MB+NC

Bài 3: 

a: Ta có: \(A=3\sqrt{3}+5\sqrt{12}-2\sqrt{27}\)

\(=3\sqrt{3}+10\sqrt{3}-6\sqrt{3}\)

\(=7\sqrt{3}\)

b: ta có: \(B=\left(\sqrt{20}-\sqrt{45}+3\sqrt{5}\right):\sqrt{5}\)

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

=2

c: ta có: \(C=\sqrt{5-\sqrt{13+\sqrt{48}}}\)

\(=\sqrt{5-2\sqrt{3}-1}\)

\(=\sqrt{3}-1\)