có sự nhầm lẫn gì đó thì phải hoặc ko

căn 31+ căn 17+ căn 3> 11

căn 31+ căn 7 +căn 3> 11

căn 31+ căn 17 +căn 3= căn 51 ko biến đổi được bỏ căn đi thì 51 >11

câu tiếp theo tương tự

Xét thấy: \(\hept{\begin{cases}31< 36\\7< 9\\3< 4\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{31}< \sqrt{36}=6\\\sqrt{7}< \sqrt{9}=3\\\sqrt{3}< \sqrt{4}=2\end{cases}}} \)

\(\Rightarrow\sqrt{31}+\sqrt{7}+\sqrt{3}< 6+3+2=11\)

Vậy: .......

căn 17 nha mn

\(M=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+1\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+1\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(=6+\sqrt{6}-11\sqrt{6}-11=-5-10\sqrt{6}\)

\(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)\)

\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+1\right)\)

\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+1\right)\)

\(M=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+1\right)\)

\(M=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+1\right)\)

\(M=\left(5\sqrt{6}-4\sqrt{6}+1-12\right)\left(\sqrt{6}+1\right)\)

\(M=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(M=6+\sqrt{6}-11\sqrt{6}-11\)

\(M=-10\sqrt{6}-5\)

Ta có: \(\sqrt{2022}-\sqrt{2021}=\dfrac{2022-2021}{\sqrt{2022}+\sqrt{2021}}=\dfrac{1}{\sqrt{2022}+\sqrt{2021}}\)

Ta có: \(\sqrt{2022}+\sqrt{2021}>1\Rightarrow\dfrac{1}{\sqrt{2022}+\sqrt{2021}}< 1\)

\(\Rightarrow\sqrt{2022}-\sqrt{2021}< 1\)

\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)

\(=6+3-7\)

\(=2\)

\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)

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

\(=\sqrt{2}\cdot2\sqrt{2}\)

\(=4\)

a) \(T=\sqrt{36}+\sqrt{9}-\sqrt{49}\)

    \(=6+3-7\)

      \(=2\)

b)  \(B=\sqrt{2\left(\sqrt{50}-3\sqrt{2}\right)}\)

         \(=\sqrt{10\sqrt{2}-6\sqrt{2}}\)

           \(=\sqrt{\left(10-6\right)\sqrt{2}}\)

           \(=\sqrt{4\sqrt{2}}\)

            \(\approx2,39\)

a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)

\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)

\(=\sqrt{518400}\)

=720

b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)

\(=9-2\sqrt{14}+2\sqrt{14}\)

=9

c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)

\(=6+2\sqrt{5}-6\)

\(=2\sqrt{5}\)

\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)

\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)

Mình sửa lại chút nha bạn

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+8\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+8\sqrt{2}}{6}\)

> nhé

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Ta có :

\(\sqrt{31}>\sqrt{30,25}=5.5\)

\(\sqrt{17}>\sqrt{16}=4\)

\(\sqrt{3}>\sqrt{2,25}=1,5\)

Từ điều trên ta có: \(\sqrt{31}+\sqrt{17}+\sqrt{3}>5,5+4+1,5=11\)

Vậy ta có : \(\sqrt{31}+\sqrt{17}+\sqrt{3}>11\)