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a)\(\sqrt{8}+3< \sqrt{9}+3=3+3=6< 6+\sqrt{2}\)
b)\(14=\sqrt{196}>\sqrt{195}=\sqrt{13.15}=\sqrt{13}.\sqrt{15}\)
c) Ta có: \(\hept{\begin{cases}\sqrt{27}>\sqrt{25}=5\\\sqrt{6}>\sqrt{4}=2\end{cases}\Rightarrow\sqrt{27}+\sqrt{6}+1>5+2+1=8}\)
Mà \(\sqrt{48}< \sqrt{49}=7< 8\)
\(\Rightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)
Tham khảo nhé~
a,\(\sqrt{12}=2\sqrt{3}=\sqrt{3}+\sqrt{3}\)
ta có \(\sqrt{5}>\sqrt{3}\)và\(\sqrt{7}>\sqrt{3}\)=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)
a, (3 + \(\sqrt{5}\))2 = 14+ \(6\sqrt{5}\) = 14 + \(\sqrt{180}\)
(\(2\sqrt{2}\) + \(\sqrt{6}\))2 = 14 + 4\(\sqrt{12}\) = 14 + \(\sqrt{192}\)
⇒ \(\left(3+\sqrt{5}\right)^2< \left(2\sqrt{2}+\sqrt{6}\right)^2\) (Do 14+\(\sqrt{180}\) < 14 + \(\sqrt{192}\))
Nên 3 + \(\sqrt{5}\) < 2\(\sqrt{2}\) + \(\sqrt{6}\) (Do \(\left(3+\sqrt{5}\right)^2< \left(2\sqrt{2}+\sqrt{6}\right)^2\))
b, c, d làm tương tự bạn nhé
\(a\)
\(\sqrt{7}+\sqrt{15}\)
\(=\sqrt{7+15}\)
\(=4,69\)
\(4,69< 7\)
\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)
\(b\)
\(\sqrt{7}+\sqrt{15}+1\)
\(=\sqrt{7+15}+1\)
\(=4,69+1\)
\(=5,69\)
\(\sqrt{45}\)
\(=6,7\)
\(5,69< 6,7\)
\(\Rightarrow\)\(\sqrt{7}+\sqrt{15}+1\)\(< \)\(\sqrt{45}\)
\(c\)
\(\frac{23-2\sqrt{19}}{3}\)
\(=\frac{22.4,53}{3}\)
\(=\frac{95,7}{3}\)
\(=31,9\)
\(\sqrt{27}\)
\(=5,19\)
\(31,9>5,19\)
\(\text{}\Rightarrow\text{}\text{}\)\(\frac{23-2\sqrt{19}}{3}\)\(>\sqrt{27}\)
\(d\)
\(\sqrt{3\sqrt{2}}\)
\(=\sqrt{3.1,41}\)
\(=\sqrt{4,23}\)
\(=2,05\)
\(\sqrt{2\sqrt{3}}\)
\(=\sqrt{2.1,73}\)
\(=\sqrt{3,46}\)
\(=1,86\)
\(2,05>1,86\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
\(Học \) \(Tốt !!!\)
a) Ta có : \(\sqrt{7}< \sqrt{9}=3;\sqrt{15}< \sqrt{16}=4\)
Do đó : \(\sqrt{7}+\sqrt{15}< 3+4=7\)
b) Ta có : \(\sqrt{17}>\sqrt{16}=4;\sqrt{5}>\sqrt{4}=2\)
\(\Rightarrow\sqrt{17}+\sqrt{5}+1>4+2+1=7\)
Lại có : \(\sqrt{45}< \sqrt{49}< 7\)
Do đó : \(\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)
c) Ta thấy : \(\sqrt{19}>\sqrt{16}=4\)
\(\Rightarrow2\sqrt{19}>2.4=8\)
\(\Rightarrow-2\sqrt{19}< -8\)
\(\Rightarrow23-2\sqrt{19}< 23-8=15\)
\(\Rightarrow\frac{23-2\sqrt{19}}{3}< 5\). Mặt khác : \(\sqrt{27}>\sqrt{25}=5\)
Nên : \(\frac{23-2\sqrt{19}}{3}< \sqrt{27}\)
d) Vì : \(18>12>0\Rightarrow\sqrt{18}>\sqrt{12}>0\)
\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}>0\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Câu b : Ta có : \(\sqrt{13}.\sqrt{15}=\sqrt{\left(14-1\right)}.\sqrt{\left(14+1\right)}=\sqrt{14}^2-1=14-1< 14\)
a: \(\left(\sqrt{18}+3\right)^2=27+18\sqrt{2}\)
\(\left(6+\sqrt{2}\right)^2=38+12\sqrt{2}\)
mà \(27+18\sqrt{2}< 38+12\sqrt{2}\)
nên \(3+\sqrt{18}< 6+\sqrt{2}\)
b: \(14=\sqrt{196}>\sqrt{195}=\sqrt{13\cdot15}\)
ta có: \(\sqrt{27}+\sqrt{6}+1=3\sqrt{3}+\sqrt{6}+1\)(1))
\(\sqrt{48}=4\sqrt{3}=3\sqrt{3}+\sqrt{3}\)(2)
ta lại có: \(\sqrt{6}>\sqrt{3}\Rightarrow\sqrt{6}+1>\sqrt{3}\) (3)
từ (1)(2)và(3)\(\Rightarrow3\sqrt{3}+\sqrt{6}+1>3\sqrt{3}+\sqrt{3}\)
\(\Leftrightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)
Ta có \(\sqrt{8}+3< \sqrt{9}+3=3+3=6\)
=> \(\sqrt{8}+3< 6\)
Ta có \(\sqrt{48}< \sqrt{49};\sqrt{35}< \sqrt{36}\)
=> \(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{46}\)
=> \(\sqrt{48}+\sqrt{35}< 13\)
=> \(\sqrt{48}< 13-\sqrt{35}\)
c) Ta có \(-\sqrt{19}< -\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{31}-\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{36}-17=6-\sqrt{17}\)
d) Ta có \(9=\sqrt{81}\Leftrightarrow\sqrt{81}>\sqrt{80}\);
\(-\sqrt{58}>-\sqrt{59}\)
=> \(\sqrt{81}-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
<=> \(9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
a) ta có \(\sqrt{27}>\sqrt{25}=5\)
\(\sqrt{6}>\sqrt{4}=2\)
Suy ra \(\sqrt{27}+\sqrt{6}+1>5+2+1=8\)
Ta có 64>48\(\Rightarrow\sqrt{64}>\sqrt{48}\Rightarrow8>\sqrt{48}\)
Vậy \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)
b) Ta có \(\sqrt{15}.\sqrt{17}=\sqrt{255}\)
Ta lại có 324>255\(\Rightarrow\sqrt{324}>\sqrt{255}\Rightarrow18>\sqrt{255}\)
Vậy \(18>\sqrt{15}.\sqrt{17}\)