a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

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

a) Có 7 = 3 + 4 = \(\sqrt{9}+\sqrt{16}\)

mà 7 < 9 => \(\sqrt{7}< \sqrt{9}\)

15 < 16 => \(\sqrt{15}< \sqrt{16}\)

=> \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

=> \(\sqrt{7}+\sqrt{15}< 7\)

Vậy \(\sqrt{7}+\sqrt{15}< 7\)

b) Có 21 > 20

=> \(\sqrt{21}>\sqrt{20}\)

=> \(\sqrt{21}-\sqrt{6}>\sqrt{20}-\sqrt{6}\) (1)

Lại có 5 < 6

=> \(\sqrt{5}< \sqrt{6}\)

=> \(-\sqrt{5}>-\sqrt{6}\)

=> \(\sqrt{21}-\sqrt{5}>\sqrt{21}-\sqrt{6}\) (2)

Từ (1) và (2) => \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

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

c) Có 27 > 25 => \(\sqrt{27}>\sqrt{25}\)

6 > 4 => \(\sqrt{6}>\sqrt{4}\)

=> \(\sqrt{27}+\sqrt{6}\) > \(\sqrt{25}+\sqrt{4}\)

=> \(\sqrt{27}+\sqrt{6}\) > 5 + 2

= >\(\sqrt{27}+\sqrt{6}+1>5+2+1\)

=> \(\sqrt{27}+\sqrt{6}+1>8\)

=> \(\sqrt{27}+\sqrt{6}+1>7\) (vì 8 > 7) (1)

Lại có 49 > 48

=> \(\sqrt{49}>\sqrt{48}\)

=> 7 > \(\sqrt{48}\) (2)

Từ (1) và (2) => \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)

Vậy \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)

\(\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{16}=3+4=7\Rightarrow\sqrt{7}+\sqrt{15}<7\)

\(\sqrt{2}+\sqrt{11}<\sqrt{3}+\sqrt{25}=\sqrt{3}+5\Rightarrow\sqrt{2}+\sqrt{11}<\sqrt{3}+5\)

\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\Rightarrow\sqrt{17}+\sqrt{26}+1>10\)

\(\sqrt{99}<\sqrt{100}=10\Rightarrow\sqrt{99}<10\)

Nên  \(\sqrt{17}+\sqrt{26}+1>10\)

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

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: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)

\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

mà \(-2\sqrt{105}>-2\sqrt{120}\)

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

b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)

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

mà \(4< 6\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)

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)

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

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Mình chọn nhầm lớp 8 chứ thật ra câu hỏi ở bên lớp 9 

a) Ta có \(5=\sqrt{25}\)

Vì \(\sqrt{25}>\sqrt{11}\) nên \(5>\sqrt{11}\)

b) Ta có \(4=\sqrt{16}\)

Vì \(\sqrt{13}< \sqrt{16}\) nên \(\sqrt{13}< 4\)

c) Ta có \(-7=-\sqrt{49}\)

Vì \(-\sqrt{49}< -\sqrt{43}\) nên \(-7< -\sqrt{43}\)

d) Ta có \(-5=-\sqrt{25}\)

Vì \(-\sqrt{21}>-\sqrt{25}\) nên \(-\sqrt{21}>-5\)