bình phương 2 vế ta có:

vế 1 bằng 50+2=52

vế 2 bằng 50+ 10+ 2 = 62

vậy (1) < (2)

vẫn k hiu hả

\(\left(\sqrt{50+2}\right)^2=52\)

\(\left(\sqrt{50}+\sqrt{2}\right)^2=62\)

\(\sqrt{50+2}=\sqrt{50}+\sqrt{2}\)

Tích nha

\(\sqrt{50+2}\)

\(=\sqrt{52}< 8\)

\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}=8\)

\(\sqrt{50+2}=\sqrt{52}< \sqrt{64}=8=7+1=\sqrt{49}+\sqrt{1}< \sqrt{50}+\sqrt{2}\)\(\sqrt{50}+\sqrt{2}\)

\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)

\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)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

a) \(\sqrt{27}+\sqrt{12}>\sqrt{25}+\sqrt{9}=5+3=8\)

\(\Rightarrow\sqrt{27}+\sqrt{12}>8\)

b) \(\sqrt{50+2}=\sqrt{52}< \sqrt{64}=8\)

\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}=7+1=8\)

=> \(\sqrt{50+2}< 8< \sqrt{50}+\sqrt{2}\)

\(\Rightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)

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 \(\sqrt{\left(1-\sqrt{50}\right)^2}>6\)

\(\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1\approx6,07>6\)

\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}>6\)

Ta có:\(\sqrt{\left(1-\sqrt{50}\right)^2}=|1-\sqrt{50}|=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)

\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2>6}\)

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) => \(\sqrt{50}+\sqrt{26}+1>13>\sqrt{168}\Rightarrow\sqrt{50}+\sqrt{26}+1>\sqrt{168}\)

\(\sqrt{50}>\sqrt{49}=7\)

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

\(\sqrt{1}=1\)

cộng vào \(VT>VP=13>\sqrt{169}>\sqrt{168}\)