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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{50+2}\)
\(=\sqrt{52}< 8\)
\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)
\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)
\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)
So sánh gì thế em, em nhập đủ đề vào hi
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{27}+\sqrt{12}>\sqrt{25}+\sqrt{9}=5+3=8\)
\(\Rightarrow\sqrt{27}+\sqrt{12}>8\)
b) \(\sqrt{50+2}=\sqrt{52}< \sqrt{64}=8\)
\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}=7+1=8\)
=> \(\sqrt{50+2}< 8< \sqrt{50}+\sqrt{2}\)
\(\Rightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:
\(\left(\sqrt{50+2}\right)^2=52\)
\(\left(\sqrt{50}+\sqrt{2}\right)^2=50+2+2\sqrt{50}\sqrt{2}\)
\(52+2\sqrt{100}=72\)
suy ra: \(\sqrt{50+2}
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(1=\sqrt{1}< \sqrt{50}\Rightarrow1-\sqrt{50}< 0\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)
Vậy \(\sqrt{\left(1-\sqrt{50}\right)^2}>6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1\approx6,07>6\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}>6\)
Ta có:\(\sqrt{\left(1-\sqrt{50}\right)^2}=|1-\sqrt{50}|=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2>6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt{50}+\sqrt{26}+1\)
Ta thấy: \(\sqrt{50}>\sqrt{49}=7,\sqrt{26}>\sqrt{25}=5\)
\(\Rightarrow A>\sqrt{49}+\sqrt{25}+1=7+5+1=13\left(1\right)\)
Ta thấy: \(\sqrt{168}< \sqrt{169}=13\left(2\right)\)
Từ (1) và (2) => \(\sqrt{50}+\sqrt{26}+1>13>\sqrt{168}\Rightarrow\sqrt{50}+\sqrt{26}+1>\sqrt{168}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{50}>\sqrt{49}=7\)
\(\sqrt{26}>\sqrt{25}=5\)
\(\sqrt{1}=1\)
cộng vào \(VT>VP=13>\sqrt{169}>\sqrt{168}\)
bình phương 2 vế ta có:
vế 1 bằng 50+2=52
vế 2 bằng 50+ 10+ 2 = 62
vậy (1) < (2)