Question

[nguyễn quỳnh anh]

So sánh

\(\sqrt{8}_{ }\)_\(\sqrt{5}\)và 1

\(\sqrt{63-27}\) và \(\sqrt{63}\)_\(\sqrt{27}\)

Answer 1

[Aquarius]

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\(\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)

Answer 2

[Huy Rio]

√8_√5< 1

√63−27 > √63_√27