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\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
a) \(\sqrt[3]{7+5\sqrt{2}}=\sqrt{2}+1\)
b) \(-6\sqrt[3]{7}=\sqrt[3]{\left(-6\right)^3\cdot7}=\sqrt[3]{-1512}\)
\(7\sqrt[3]{-6}=\sqrt[3]{7^3\cdot\left(-6\right)}=\sqrt[3]{-2058}\)
mà -1512>-2058
nên \(-6\sqrt[3]{7}>7\cdot\sqrt[3]{-6}\)
a) 7 và \(\sqrt{37}+1\)
=7 và 7,08
=>......
b) \(\sqrt{17}-\sqrt{50}-1\)và \(\sqrt{99}\)
=-3,95 và 9,95
=>.....
\(\sqrt{3\sqrt{3}}=\sqrt{3.\sqrt{\dfrac{432}{144}}}< \sqrt{3\sqrt{\dfrac{625}{144}}}=\sqrt{3.\dfrac{25}{12}}=\dfrac{5}{2}\)
\(\sqrt{3}+1=\sqrt{\dfrac{12}{4}}+1>\sqrt{\dfrac{9}{4}}+1=\dfrac{3}{2}+1=\dfrac{5}{2}\)
\(\Rightarrow\sqrt{3}+1>\sqrt{3\sqrt{3}}\)
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)
\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
Do đó: A=B
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}=\left|\sqrt{5}+1\right|-\sqrt{5}=1\)
\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
--> Bằng nhau
Ta có :
\(\sqrt[3]{26}\approx3\); \(\sqrt[3]{7}\approx2\)
\(\Leftrightarrow\sqrt[3]{26}+\sqrt[3]{7}\approx5\)
Mà: \(\sqrt[3]{126}+1\approx6\)
\(\Leftrightarrow\sqrt[3]{26}+\sqrt[3]{7}< 1+\sqrt[3]{126}\)
giúp với ạ mik k cho