b) Ta có: \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{5+35}{7+49}=\frac{40}{56}=\frac{5}{7}\) (1)

Lại có: \(\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\frac{5-35}{7-49}=\frac{-30}{-42}=\frac{5}{7}\) (2)

Từ biểu thức (1) và biểu thức (2)

=> \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)

\(\frac{\sqrt{21} - \sqrt{13}}{35 - 2\sqrt{273}} = \frac{\sqrt{21} - \sqrt{13}}{(\sqrt{21} - \sqrt{13})^2} = \frac{1}{\sqrt{21} - \sqrt{13}}\)

\(=\frac{\sqrt{21}+\sqrt{13}}{8}>\frac{\sqrt{16}+\sqrt9}{8}=\frac{4+3}{8}=\frac78\)

\(\frac{\sqrt{10} - \sqrt{5}}{16 - 10\sqrt{2}}=\frac{\sqrt{5}(\sqrt{2} - 1)}{16 - 10\sqrt{2}}=\frac{\sqrt{5}}{6\sqrt{2} - 4}=\frac{\sqrt5}{2\left(3\sqrt2-2\right)}\)

\(\frac{\sqrt5}{2\left(3\sqrt2-2\right)}-\frac18=\frac{8\sqrt5-6\sqrt2+4}{16\left(3\sqrt2-2\right)}=\frac{\sqrt{320}-\sqrt{72}+4}{16\left(3\sqrt2-2\right)}>\frac{17-8+4}{16\left(3\sqrt2-2\right)}=\frac{13}{16\left(3\sqrt2-2\right)}>5>\frac18\)

Do đó: \(\frac{\sqrt{21}-\sqrt{13}}{35-2\sqrt{273}}+\frac{\sqrt{10}-\sqrt5}{16-10\sqrt2}\) >1(ĐPCM)

Đặt \(A=\frac{\sqrt5+1}{5\cdot\sqrt{10-2\sqrt5}};B=\frac{\sqrt3}{6}\)

\(A^2 = \frac{(\sqrt{5}+1)^2}{25(10-2\sqrt{5})} = \frac{5 + 2\sqrt{5} + 1}{25(10-2\sqrt{5})} = \frac{6 + 2\sqrt{5}}{25(10-2\sqrt{5})}\)

\(=\frac{3 + \sqrt{5}}{25(5-\sqrt{5})}=\frac{(3+\sqrt{5})(5+\sqrt{5})}{25(5^2 - 5)}\)

\(=\frac{15 + 3\sqrt{5} + 5\sqrt{5} + 5}{25 \cdot20}=\frac{20 + 8\sqrt{5}}{500}=\frac{5 + 2\sqrt{5}}{125}\)

=>\(A^2\) ≃0,075

\(B^2=\left(\frac{\sqrt3}{6}\right)^2=\frac{3}{36}=\frac{1}{12}\) ≃0,083

=>\(A^2

=>A<B

\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

a) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)

\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

Do đó: A=B

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

\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

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