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a)\(1+\sqrt{3}>1+\sqrt{1}=1+1=2\)
Vậy \(1+\sqrt{3}>2\)
c) \(\sqrt{3}-1< \sqrt{4}-1=2-1=1\)
Vậy \(\sqrt{3}-1< 1\)
e) \(\sqrt{2}+\sqrt{5}< \sqrt{16}+\sqrt{16}=4+4=8\)
Vậy \(\sqrt{2}+\sqrt{5}< 8\)
\(\left(\sqrt{8}+\sqrt{11}\right)^2=8+2\sqrt{8}.\sqrt{11}+11=19+\sqrt{352}\)\(< 19+\sqrt{361}=19+19=38=\left(\sqrt{38}\right)^2\)
\(\Leftrightarrow\left(\sqrt{8}+\sqrt{11}\right)^2< \left(\sqrt{38}\right)^2\Rightarrow\sqrt{8}+\sqrt{11}< \sqrt{38}\)
Anh ko hiểu chỗ nào thì hỏi em nhé!
\(\frac{1+\sqrt{3}}{\sqrt{3}-1}=\frac{\left(1+\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=2+\sqrt{3}\)
\(\frac{2}{\sqrt{2}-1}=\frac{2\sqrt{2}+2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=\sqrt{8}+2\)
\(\Rightarrow\frac{2}{\sqrt{2}-1}>\frac{1+\sqrt{3}}{\sqrt{3}-1}\)
\(A=\sqrt{11+\sqrt{96}}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=2\sqrt{2}+\sqrt{3}\)
\(B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=1+\sqrt{2}+\sqrt{3}\)
Xét : \(A-B=2\sqrt{2}+\sqrt{3}-\left(1+\sqrt{2}+\sqrt{3}\right)=\sqrt{2}-1>0\)
\(\Rightarrow A>B\)
a: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)
\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)
mà \(2\sqrt{22}< 15+10\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)
b: \(\left(\sqrt{8}+\sqrt{11}\right)^2=19+2\cdot\sqrt{88}=19+\sqrt{352}\)
\(\left(\sqrt{38}\right)^2=19+19=19+\sqrt{361}\)
mà 352<361
nên \(\sqrt{8}+\sqrt{11}< \sqrt{38}\)