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>\(\sqrt{24}+\sqrt{3}\)
\(7=5+2=\sqrt{25}+\sqrt{4}>\sqrt{24}+\sqrt{3}\)
\(\left(\sqrt{50+2}\right)^2=\left|50+2\right|=50+2\\ \left(\sqrt{50}+\sqrt{2}\right)^2=50+20+2>50+2=\left(\sqrt{50+2}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{50+2}\right)^2}< \sqrt{\left(\sqrt{50}+\sqrt{2}\right)^2}\\ \Leftrightarrow\left|\sqrt{50+2}\right|< \left|\sqrt{50}+\sqrt{2}\right|\\ \Leftrightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)
\(\left(\sqrt{63-27}\right)^2=\left|63-27\right|=63-27\\ \left(\sqrt{63}-\sqrt{27}\right)^2=63-2\sqrt{90}+27>63-20+27=63-7>63-27=\left(\sqrt{63-27}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{63-27}\right)^2}< \sqrt{\left(\sqrt{63}-\sqrt{27}\right)^2}\\ \Leftrightarrow\left|\sqrt{63-27}\right|< \left|\sqrt{63}-\sqrt{27}\right|\\ \Leftrightarrow\sqrt{63-27}< \sqrt{63}-\sqrt{27}\)
b, Ta có:
\(\sqrt{50+2}=\sqrt{52}=7,21\\
\sqrt{50}+\sqrt{2}=7,07+1,41=8,48
\)
vì 7,21<8,48 nên \(\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)
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c) Ta có:
\(\left(\sqrt{63-27}\right)^2=63-27=63+27-54\)
\(\left(\sqrt{63}-\sqrt{27}\right)^2=\left(\sqrt{63}\right)^2-2\sqrt{63}\sqrt{27}+\left(\sqrt{27}\right)^2=63+27-2\sqrt{63}\sqrt{27}\)
Ta lại có: \(\left\{{}\begin{matrix}\sqrt{63}>\sqrt{49}=7\\\sqrt{27}>\sqrt{25}=5\end{matrix}\right.\)
\(\Rightarrow2\sqrt{63}\sqrt{27}>70>54\)
Vì \(2\sqrt{63}\sqrt{27}>54\) nên \(63+27-54>63+27-2\sqrt{63}\sqrt{27}\)
\(\Rightarrow\left(\sqrt{63-27}\right)^2>\left(\sqrt{63}-\sqrt{27}\right)^2\)
\(\Rightarrow\sqrt{63-27}>\sqrt{63}-\sqrt{27}\)
Vậy...
c) Ta có : \(\sqrt{63}-\sqrt{27}< \sqrt{64}-\sqrt{27}=8-3=5< 6=\sqrt{36}=\sqrt{63-27}\)
Vậy \(\sqrt{63-27}>\sqrt{63}-\sqrt{27}\)
