\(\sqrt{27}>\sqrt{25}=5.\)
\(\sqrt{26}>\sqrt{25}=5.\)
\(\sqrt{27}+\sqrt{26}+1>5+5+1=11.\)
\(\sqrt{99}< \sqrt{100}=10\)
\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
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 \(:\)\(\sqrt{27}+\sqrt{26}>5+5=10>\sqrt{99}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
\(\sqrt{27}\)>\(\sqrt{25}\)=5
\(\sqrt{26}\)>\(\sqrt{25}\)=5
\(\sqrt{27+\sqrt{26}}\)\(+>5+5+1=11\)
\(\sqrt{99}\)\(< \sqrt{100}\)=\(10\)
\(\sqrt{27}\)\(+\)\(\sqrt{26}\)\(+1\)\(>\)\(\sqrt{99}\)
k mik nha
ta có\(\sqrt{27}>\sqrt{25}=5\)
\(\sqrt{26}>\sqrt{25}=5\)
Cộng từng vế của bdt ta được
\(\sqrt{27}+\sqrt{26}>5+5\)
=> \(\sqrt{27}+\sqrt{26}+1>5+5+1=11\left(1\right)\)
Lại có \(\sqrt{99}< \sqrt{100}=10\left(2\right)\)
từ (1) và (2)=>\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
vậy \(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
Ta có: \(\hept{\begin{cases}\sqrt{27}>\sqrt{25}\\\sqrt{26}>\sqrt{25}\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{27}>5\\\sqrt{26}>5\end{cases}}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}>10\Rightarrow\sqrt{27}+\sqrt{26}+1>11\left(1\right)\)
Ta lại có :\(\sqrt{99}< \sqrt{121}=11\left(2\right)\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow1+\sqrt{27}+\sqrt{26}>\sqrt{99}\)
\(\sqrt{27}>\sqrt{25}=5\)
\(\sqrt{26}>\sqrt{25}=5\)
\(\sqrt{27}+\sqrt{26}+1>5+5+1=11\)
\(\sqrt{99}< \sqrt{100}=10\)
\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
