\(\sqrt{27}>\sqrt{25}=5.\)

\(\sqrt{26}>\sqrt{25}=5.\)

\(\sqrt{27}+\sqrt{26}+1>5+5+1=11.\)

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

\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

ta có : \(\sqrt{27}+\sqrt{26}+1\approx11,29\)

                \(\sqrt{99}\approx9,94\)

\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

\(\sqrt{27}>\sqrt{25}=5\)(1)

\(\sqrt{26}>\sqrt{25}=5\)(2)

\(\sqrt{99}< \sqrt{100}=10\)(3)

Từ (1) và (2) và (3) ta có \(:\)\(\sqrt{27}+\sqrt{26}>5+5=10>\sqrt{99}\)

                        \(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

\(\sqrt{37}>6\)

\(-\sqrt{14}>-\sqrt{15}\)

=> \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

\(\sqrt{27}-\sqrt{14}>6-\sqrt{15}\)

Ta có:

\(\sqrt[3]{7}< \sqrt[3]{8}=2\) và \(\sqrt{15}< \sqrt{16}=4\), suy ra \(\sqrt[3]{7}+\sqrt{15}< 6\).

\(\sqrt{10}>\sqrt{9}=3\) và \(\sqrt[3]{28}>\sqrt[3]{27}=3\), suy ra \(\sqrt{10}+\sqrt[3]{28}>6\).

Vậy \(\sqrt[3]{7}+\sqrt{15}< \sqrt{10}+\sqrt[3]{28}\).

a) Ta có:

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

Mà: \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\)

\(\Rightarrow1+\sqrt{1}< \sqrt{2}+1\)

\(\Rightarrow2< \sqrt{2}+1\)

b) Ta có:

\(1=2-1=\sqrt{4}-1\)

Mà: \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\)

\(\Rightarrow\sqrt{4}-1>\sqrt{3}-1\)

\(\Rightarrow1>\sqrt{3}-1\)

c) Ta có:

\(10=2\cdot5=2\sqrt{25}\)

Mà: \(25< 31\Rightarrow\sqrt{25}< \sqrt{31}\)

\(\Rightarrow2\sqrt{25}< 2\sqrt{31}\)

\(\Rightarrow10< 2\sqrt{31}\)

d) Ta có:

\(-12=-3\cdot4=-3\sqrt{16}\)

Mà: \(16>11\Rightarrow\sqrt{16}>\sqrt{11}\)

\(\Rightarrow-3\sqrt{16}< -3\sqrt{11}\)

\(\Rightarrow-12< -3\sqrt{11}\)

Lời giải:
\(\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{11}}> \frac{1}{\sqrt{4}}+\frac{1}{\sqrt{9}}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)

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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 \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)

Giả sử 

\(\frac{23-2\sqrt{19}}{3}< \sqrt{27}\)

\(\Leftrightarrow23-2\sqrt{29}< 3\sqrt{27}\)

\(\Leftrightarrow23< 3\sqrt{27}+2\sqrt{19}\)

Ta có

\(3\sqrt{27}+2\sqrt{19}>3\sqrt{25}+2\sqrt{16}=23\)

Vậy giả sử là đúng