\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)

\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}\Rightarrow A=60^0\)

\(\Rightarrow C=180^0-\left(A+B\right)=75^0\)

\(h_a=\frac{bc.sinA}{a}=\frac{2.\left(\sqrt{3}+1\right)sin60^0}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{2}}{2}\)

Bài 1:

a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)

b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\) 

mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)

Bài 2:

a. \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\dfrac{-\sqrt{2}}{-1}=\sqrt{2}\)

b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}\)

\(=21\)

c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=32\)

a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)

b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)

d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=t\sqrt{3}\\b=t\sqrt{2}\\c=\frac{t\left(\sqrt{6}-\sqrt{2}\right)}{2}\end{matrix}\right.\)

\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{2t^2+\left(2-\sqrt{3}\right)t^2-3t^2}{t^2.\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}=-\frac{1}{2}\)

\(\Rightarrow A=120^0\)

\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)

\(\Rightarrow C=180^0-\left(A+B\right)=15^0\)

\(R=\frac{a}{2sinA}=\frac{2\sqrt{3}}{2sin120^0}=2\)

Bài 1:

a: \(\sqrt{\left(2\sqrt6-4\right)^2}+\sqrt{15-6\sqrt6}\)

\(=\left|2\sqrt6-4\right|+\sqrt{\left(3-\sqrt6\right)^2}\)

\(=2\sqrt6-4+3-\sqrt6=-1+\sqrt6\)

b: \(\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{19+2\cdot\sqrt{18}}\)

\(=\left|3-2\sqrt2\right|+\sqrt{\left(3\sqrt2+1\right)^2}\)

\(=3-2\sqrt2+3\sqrt2+1=4+\sqrt2\)

c: \(\sqrt{9+4\sqrt5}-\sqrt{\left(1-\sqrt5\right)^2}\)

\(=\sqrt{\left(\sqrt5+2\right)^2}-\left|1-\sqrt5\right|\)

\(=\sqrt5+2-\left(\sqrt5-1\right)=3\)

Bài 2:

a: \(\frac{1}{\sqrt7+3}+\frac{1}{\sqrt7-3}=\frac{\sqrt7-3+\sqrt7+3}{\left(\sqrt7-3\right)\left(\sqrt7+3\right)}=\frac{2\sqrt7}{7-9}=-\sqrt7\)

b: \(\frac{3}{\sqrt2-1}+\frac{\sqrt6+\sqrt2}{\sqrt3+1}\)

\(=\frac{3\left(\sqrt2+1\right)}{\left(\sqrt2-1\right)\left(\sqrt2+1\right)}+\frac{\sqrt2\left(\sqrt3+1\right)}{\sqrt3+1}\)

\(=3\left(\sqrt2+1\right)+\sqrt2=4\sqrt2+3\)

c: \(\frac{1}{7+4\sqrt3}+\frac{1}{7-4\sqrt3}=\frac{7-4\sqrt3+7+4\sqrt3}{\left(7+4\sqrt3\right)\left(7-4\sqrt3\right)}=\frac{14}{49-48}=14\)

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

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

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

b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)

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

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

\(=2\sqrt{3}\)

`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`

`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`

`=sqrt{3-2}=1`

`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`

`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`

`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`

`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`

`c)C=3-sqrt{3-sqrt5}`

`=3-sqrt{(6-2sqrt5)/2}`

`=3-sqrt{(sqrt5-1)^2/2}`

`=3-(sqrt5-1)/sqrt2`

`=3-(sqrt{10}-sqrt2)/2`

`=(6-sqrt{10}+sqrt2)/2`

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

b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)

c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)

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