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a, ta có:
\(\sqrt{24}=4,89\\ \sqrt{3}=1,73\)
\(\Rightarrow\sqrt{24}+\sqrt{3}=4,89+1,73=6,62\)
vì 7>6,62 nên 7>\(\sqrt{24}+\sqrt{3}\)
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Ta có:
\(\sqrt{63-27}=\sqrt{36}=6\)
\(\sqrt{63}-\sqrt{27}
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Ta có\(8< 16\Rightarrow\sqrt{8}< \sqrt{16}=4\)
và \(5< 9\Rightarrow\sqrt{5}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{8}-\sqrt{5}< \sqrt{16}-\sqrt{9}=4-3=1\)
Vậy \(\sqrt{8}-\sqrt{5}< 1\)
Ta có \(\sqrt{63-27}=\sqrt{36}=6\)
lại có\(63< 64\Rightarrow\sqrt{63}< \sqrt{64}=8\)và \(27>4\Rightarrow\sqrt{27}>\sqrt{4}=2\)
\(\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{64}-\sqrt{4}=8-2=6\)
mà\(\sqrt{63-27}=6\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)
Vậy\(\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)
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a, Ta có: 1= \(\sqrt{9}-\sqrt{4}\)
Nx: \(\begin{cases} 0<8<9\\ 5>4>0 \end{cases}\)
=>\(\begin{cases} \sqrt{8}<\sqrt{9}\\ \sqrt{5}>\sqrt{4} \end{cases}\)
=>\(\sqrt{8}-\sqrt{5}<\sqrt{9}-\sqrt{4}\)
=>\(\sqrt{8}-\sqrt{5}<1\)
b,Ta có: \(\sqrt{63-27}=\sqrt{36}=6\)
\(\sqrt{63}-\sqrt{27}<\sqrt{64}-\sqrt{25}\\ =>\sqrt{63}-\sqrt{27}<3\\ =>\sqrt{63}-\sqrt{27}<6(vì 3<6)\\ =>\sqrt{63}-\sqrt{27}<\sqrt{63-27} \)
![](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) \(\sqrt{50+2}=\sqrt{52}< \sqrt{64}=8\)
\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}=7+1=8\)
=> \(\sqrt{50+2}< 8< \sqrt{50}+\sqrt{2}\)
\(\Rightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)