Đề bài:
So sánh \(\sqrt{17}+\sqrt{26}\)và \(\sqrt{99}\)

\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

HAY  \(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

\(\sqrt{2019}-\sqrt{2017}\approx0,022\) \(< \sqrt{19}-\sqrt{17}\approx0,23\)

chắc vậy :))

Ta có:\(\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}=4+5=9\)

Hay \(\sqrt{17}+\sqrt{26}>9\)

= \(\sqrt{17}+\sqrt{26}\)và 9 

\(\sqrt{17}=4,123105626\)

\(\sqrt{26}=5,099019514\)

\(=4,123105626+5,099019514=9,222,25139\)

Vậy \(\sqrt{17}+\sqrt{26}>9\)

\(\sqrt{17} + \sqrt{26} + 1 \approx 10.222\) và \(\sqrt{99} \approx 9.949\), nên ta có:

\(\sqrt{17} + \sqrt{26} + 1 > \sqrt{99}\)

ta có: \(\sqrt{17}>\sqrt{16}=4,\sqrt{26}>\sqrt{25}=5\)
và \(\sqrt{99}<\sqrt{100}=10\)
nên :\(\sqrt{17}+\sqrt{26}+1>4+5+1=10>\sqrt{99}\)
Vậy : \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

cmr là cái j

Lê Thanh Thùy Ngân 

cmr là chứng minh rằng bạn nhé 

c) Bình phương hai vế ta được 2015+2017+2\(\sqrt{2015\times2017}\) và 4\(\times\)2016

Ta có 2015 + 2017 + 2\(\sqrt{2015\times2017}\)

= (2016-1) + (2016+1) + 2\(\sqrt{2015\times2017}\)

= 2016 + 2016 + 1 - 1 + 2\(\sqrt{2015\times2017}\)

= 2\(\times\)2016 + 2\(\sqrt{2015\times2017}\) (1)

ta thấy 2015 \(\times\) 2017 =(2016-1) \(\times\) (2016+1)= 2016² - 1

nên (1) \(\Leftrightarrow\)2\(\times\)2016 + 2\(\sqrt{2016^2-1}\)

Ta có 4\(\times\)2016=2\(\times\)2016 + 2\(\times\)2016=2\(\times\)2016 + 2\(\sqrt{2016^2}\)

Vì 2016²-1 < 2016² nên 2\(\sqrt{2016^2-1}\) < 2\(\sqrt{2016^2}\)

\(\Leftrightarrow\) 2\(\times\)2016 + 2\(\sqrt{2016^2-1}\) < 2\(\times\)2016 + 2\(\sqrt{2016^2}\)

\(\Leftrightarrow\)2015+2017+2\(\sqrt{2015\times2017}\) < 4\(\times\)2016

Hay \(\sqrt{2015}+\sqrt{2017}\) < \(2\sqrt{2016}\)

a) Bình phương hai vế ta được 5+7+\(2\sqrt{5\times7}\) và 13.

Ta có 5+7+\(2\sqrt{5\times7}\) =12+\(2\sqrt{35}\)

13=12+1=12+\(2\times\frac{1}{2}\) =12+\(2\sqrt{\frac{1}{4}}\)

Vì 35 > \(\frac{1}{4}\) nên \(\sqrt{35}\) > \(\sqrt{\frac{1}{4}}\) \(\Leftrightarrow\)2\(\sqrt{35}\) > \(2\sqrt{\frac{1}{4}}\) \(\Leftrightarrow\)12+2\(\sqrt{35}\) > 12+\(2\sqrt{\frac{1}{4}}\)

Hay\(\sqrt{5}\)+\(\sqrt{7}\) > \(\sqrt{13}\)

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HK tot

Giả sử \(8>\sqrt{15}+\sqrt{17}\)

\(\Leftrightarrow64>32+2\sqrt{15×17}\)

\(\Leftrightarrow16>\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\left(dung\right)\)

Vậy \(8>\sqrt{15}+\sqrt{17}\)

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