\(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}}\)

\(=\sqrt{\dfrac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{11}+\sqrt{\dfrac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{13}}}\)

\(=\sqrt{\dfrac{22-11\sqrt{3}}{11}+\sqrt{\dfrac{13\sqrt{3}+26}{13}}}\)

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

Còn nếu không có dấu căn ở chỗ kia thì kết quả sẽ là :

\(\sqrt{2-\sqrt{3}+2+\sqrt{3}}=\sqrt{4}=2\)

 Cần ít nhất số chuồng để nhốt hết số thỏ đó là

      34:6=5 dư 4 con thỏ 

           Đáp số:5 chuồng

góc IDN = góc IDM ( ID phân giác góc MDN )

=> góc IDN = EDK = IDM

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

1. a)\(2\&\sqrt{5}\)

\(2=\sqrt{4}\)

=> \(2< \sqrt{5}\)

b)\(5\&\sqrt{23}\)

\(5=\sqrt{25}\)

=> \(5>\sqrt{23}\)

c) \(\sqrt{23}+\sqrt{13}\&\sqrt{83}\)

\(\left(\sqrt{23}+\sqrt{13}\right)^2=36+2\sqrt{229}\)

\(\left(\sqrt{83}\right)^2=83\)

\(\Rightarrow36+2\sqrt{299}< 83\)

=> \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)

2. a) \(\sqrt{x}=5;x\ge0\)

=> x = 25

b) \(3\sqrt{x}=6;x\ge0\)

=> x = 4

c) trùng

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

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

\(\Delta ABC\) vuông tại A

\(BC=\dfrac{AB^2}{BH}=\dfrac{15^2}{9}=25\)(cm)

\(AC=\sqrt{25^2-15^2}=20\)(cm)

b) ta có: \(BM=\dfrac{BC}{2}=\dfrac{25}{2}=12,5\)(cm)

=> MH = BM - BH = 12,5 - 9= 3,5 (cm)

HC = BC - HB = 12(cm)

\(AH=\sqrt{9.16}=12\)(cm)

\(S_{\Delta_{AHM}}=\dfrac{1}{2}.12.3,5=21\left(cm^2\right)\)

Gọi A₂O₃ là oxit cần tìm

\(n_{HCl}=\dfrac{5,475}{36,5}=0,15\left(mol\right)\)

\(A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\)

2M_A + 48.. 6

4g ........ 0,15mol

=> M_A = 56

A là Fe

cthc: Fe₂O₃