19) \(\sqrt{12-\sqrt{44}}=\sqrt{12-2\sqrt{11}}=\sqrt{\left(\sqrt{11}\right)^2-2\sqrt{11}.1+1^2}\)

\(=\sqrt{\left(\sqrt{11}-1\right)^2}=\sqrt{11}-1\)

20) \(\sqrt{5-\sqrt{24}}=\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{2}.\sqrt{3}+\left(\sqrt{2}\right)^2}\)

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

17) \(\sqrt{6-\sqrt{20}}=\sqrt{6-2\sqrt{5}}=\sqrt{1^2-2\sqrt{5}.1+\left(\sqrt{5}\right)^2}\)

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

18) \(\sqrt{8+\sqrt{28}}=\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}\right)^2+2\sqrt{7}+1^2}\)

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

13) \(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=2+\sqrt{5}\)

14) \(\sqrt{12+6\sqrt{3}}=\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3+\sqrt{3}\right)^2}=3+\sqrt{3}\)

15) \(\sqrt{27-10\sqrt{2}}=\sqrt{5^2-2.5.\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(5-\sqrt{2}\right)^2}=5-\sqrt{2}\)

16) \(\sqrt{18-6\sqrt{5}}=\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{15}-\sqrt{3}\right)^2}=\sqrt{15}-\sqrt{3}\)

17) \(\sqrt{21+4\sqrt{5}}=\sqrt{21+2\sqrt{20}}=\sqrt{\left(\sqrt{20}\right)^2+2\sqrt{20}.1+1^2}\)

\(=\sqrt{\left(\sqrt{20}+1\right)^2}=\sqrt{20}+1\)

18) \(\sqrt{28-6\sqrt{3}}=\sqrt{28-2\sqrt{27}}=\sqrt{\left(\sqrt{27}\right)^2-2\sqrt{27}.1+1^2}\)

\(=\sqrt{\left(\sqrt{27}-1\right)^2}=\sqrt{27}-1\)

19) \(\sqrt{15-10\sqrt{2}}=\sqrt{15-2\sqrt{50}}=\sqrt[]{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{10}+\left(\sqrt{10}\right)^2}\)

\(=\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}=\sqrt{10}-\sqrt{5}\)

20) \(\sqrt{46-6\sqrt{5}}=\sqrt{46-2\sqrt{45}}=\sqrt{\left(\sqrt{45}\right)^2-2.\sqrt{45}.1+1^2}\)

\(=\sqrt{\left(\sqrt{45}-1\right)^2}=\sqrt{45}-1\)

27) \(\sqrt{5+\sqrt{21}}=\sqrt{\dfrac{10+2\sqrt{21}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt[]{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}+\sqrt{7}\right)^2}{2}}=\dfrac{\sqrt{3}+\sqrt{7}}{\sqrt{2}}\)

28) \(\sqrt{6-\sqrt{35}}=\sqrt{\dfrac{12-2\sqrt{35}}{2}}=\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{5}.\sqrt{7}+\left(\sqrt{5}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}-\sqrt{5}\right)^2}{2}}=\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\)

29) \(\sqrt{7+\sqrt{40}}=\sqrt{\dfrac{14+2\sqrt{40}}{2}}=\sqrt{\dfrac{\left(\sqrt{4}\right)^2+2\sqrt{4}.\sqrt{10}+\left(\sqrt{10}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{4}+\sqrt{10}\right)^2}{2}}=\dfrac{\sqrt{4}+\sqrt{10}}{2}\)

30) \(\sqrt{8+\sqrt{15}}=\sqrt{\dfrac{16+2\sqrt{15}}{2}}=\sqrt{\dfrac{\left(\sqrt{15}\right)^2+2\sqrt{15}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{15}+1\right)^2}{2}}=\dfrac{\sqrt{15}+1}{\sqrt{2}}\)

31) \(\sqrt{9-\sqrt{77}}=\sqrt{\dfrac{18-2\sqrt{77}}{2}}=\sqrt{\dfrac{\left(\sqrt{11}\right)^2-2\sqrt{11}.\sqrt{7}+\left(\sqrt{7}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}-\sqrt{7}\right)^2}{2}}=\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{2}}\)

32) \(\sqrt{10+\sqrt{99}}=\sqrt{\dfrac{20+2\sqrt{99}}{2}}=\sqrt{\dfrac{\left(\sqrt{9}\right)^2+2\sqrt{9}.\sqrt{11}+\left(\sqrt{11}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{9}+\sqrt{11}\right)^2}{2}}=\dfrac{\sqrt{9}+\sqrt{11}}{\sqrt{2}}\)

21) \(\sqrt{8-\sqrt{60}}=\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{15}-\left(\sqrt{5}\right)^2}\)

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

22) \(\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

23) \(\sqrt{9+\sqrt{56}}=\sqrt{9+2\sqrt{14}}=\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.\sqrt{7}+\left(\sqrt{7}\right)^2}\)

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

24) \(\sqrt{7+\sqrt{24}}=\sqrt{7+2\sqrt{6}}=\sqrt{1^2+2\sqrt{6}.1+\left(\sqrt{6}\right)^2}=\sqrt{\left(1+\sqrt{6}\right)^2}\)

\(=1+\sqrt{6}\)

25) \(\sqrt{3-\sqrt{5}}=\sqrt{\dfrac{6-2\sqrt{5}}{2}}=\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}{2}}=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}\)

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

26) \(\sqrt{4+\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2\sqrt{7}+1}{2}}=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}\)

cách 2

Áp dụng BĐT Cosi cho 2 cặp số dương là  \(\dfrac{1}{x};\dfrac{9}{16}x\) và \(\dfrac{1}{4y};\dfrac{9}{16}y\) , ta có:

\(\dfrac{1}{x}+\dfrac{9}{16}x\ge2\sqrt{\dfrac{1}{x}.\dfrac{9}{16}x}=2.\dfrac{3}{4}=\dfrac{3}{2}\)

\(\dfrac{1}{4y}+\dfrac{9}{16}y\ge2\sqrt{\dfrac{1}{4y}.\dfrac{9}{16}y}=2.\dfrac{3}{8}=\dfrac{3}{4}\)

Cộng vế theo vế ta được: \(\dfrac{1}{x}+\dfrac{1}{4y}+\dfrac{9}{16}\left(x+y\right)\ge\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)

\(\Leftrightarrow A+\dfrac{9}{16}.2\ge\dfrac{9}{4}\Leftrightarrow A\ge\dfrac{9}{4}-\dfrac{9}{8}=\dfrac{9}{8}\)

Dấu bằng xảy ra \(\Leftrightarrow\left(x,y\right)=\left(\dfrac{4}{3};\dfrac{2}{3}\right)\)

mik đánh 

đoạn cuối là \(\Leftrightarrow\left(x-3y-2\right)\left(x+y-1\right)=4\)

\(2x-7\sqrt{x}+3\)

\(=\left(2x-\sqrt{x}\right)-\left(6\sqrt{x}-3\right)\)

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

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