mk mới hok lớp  6 nên chưa bt cái này là cái gì

\(=\sqrt{30}+2\sqrt{6}\)

Đặt \(C=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

                  \(=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}=\frac{6}{6}=1\)

\(\frac{C}{\sqrt{2}}=1\Rightarrow C=\sqrt{2}\)

Vậy ....

a)  \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)

b)  \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)

d)  \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)

Giả sử tam giác ABC vuông tại A, có góc C = 60⁰. AC = 25

A B C 60 25

\(cosC=cos60^0=\frac{AC}{BC}=\frac{1}{2}\)

\(\Rightarrow\)\(BC=2AC=50\)

\(tanC=tan60^0=\frac{AB}{AC}=\sqrt{3}\)

\(\Rightarrow\)\(AB=\sqrt{3}.AC=25\sqrt{3}\)

Hình vẽ:

60 25 A C B

Giả sử tam giác ABC vuông tại A có góc C bằng 60⁰,  AC bằng 25 (như hình vẽ)

cosC = cos60⁰ = \(\frac{AC}{BC}\)= \(\frac{1}{2}\)

=> BC = 2AC = 50

tanC = tan60⁰ = \(\frac{AB}{AC}\)= \(\sqrt{3}\)

=> AB = \(\sqrt{3}\).AC = 25 \(\sqrt{3}\)

a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )

b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)

                                                         \(=6-3b\) (vì b < 2 )

b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\) 

                                         \(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)