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a/ \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2.3}=5+2\sqrt{6}=5+\sqrt{24}\)
\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)
Vì \(\sqrt{24}< \sqrt{25}\)
=>\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)
b/\(\left(\sqrt{3}+2\right)^2=3+4+4\sqrt{3}=7+4\sqrt{3}\)
\(\left(\sqrt{2}+\sqrt{16}\right)^2=2+16+2\sqrt{2.16}=18+4\sqrt{8}\)
=> \(\sqrt{3}+2< \sqrt{2}+\sqrt{16}\)
c/ \(16=\sqrt{16^2}\)
\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)
=> \(16>\sqrt{15}.\sqrt{17}\)
d/\(8^2=64=32+32=32+2\sqrt{256}\)
\(\left(\sqrt{15}+\sqrt{17}\right)^2=15+17+2\sqrt{15.17}=32+2\sqrt{255}\)
=> \(8>\sqrt{15}+\sqrt{17}\)
a) Có \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)
\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)
=> \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
b) \(\dfrac{13-2\sqrt{3}}{6}>\dfrac{13-2\sqrt{4}}{6}=1,5\)
Mặt khác (1,5)2 = 2,25 ; \(\left(\sqrt{2}\right)^2=2\)
=> 1,5 > \(\sqrt{2}\) , do đó \(\dfrac{13-2\sqrt{3}}{6}>\sqrt{2}\)
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:
\(6\sqrt{5}=\sqrt{5\cdot36}=\sqrt{180}\)
\(5\sqrt{6}=\sqrt{6\cdot25}=\sqrt{200}\)
Mà \(\sqrt{180}< \sqrt{200}\)
Vậy: \(6\sqrt{5}< 5\sqrt{6}\)
x) Ta có: \(\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\)
Công hai vế của BĐT cho 3:
Suy ra: \(\sqrt{8}+3< 3+3=6\)
Vậy: \(\sqrt{8}+3< 6\)
b) Ta có:
\(\sqrt{2\sqrt{3}}=\sqrt[4]{12}\)
Tương tự: \(\sqrt{3\sqrt{2}}=\sqrt[4]{18}\)
Mà \(\sqrt[4]{18}>\sqrt[4]{12}\)
Vậy.....
d) Ta có:
\(2\sqrt{5}-5=\sqrt{5}+\sqrt{5}-5=\left(\sqrt{5}-2\right)+\left(\sqrt{5}-3\right)>\sqrt{5}-3\)
Vậy ......
e) Ta có:
\(\sqrt{2}-2=\frac{3\sqrt{2}-6}{3}\)
\(\sqrt{3}-3=\frac{2\sqrt{3}-6}{2}\)
Mà \(3\sqrt{2}>2\sqrt{3}\)
Vậy .....
f) ........... Đang thinking
a) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)
\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)
\(\Rightarrow\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
b) \(\frac{13-2\sqrt{3}}{6}>\frac{13-2\sqrt{4}}{6}=1,5\)
mà 1,52 = 2,25 ; \(\sqrt{2}^2=2\)
\(\Rightarrow1,5>\sqrt{2}\)hay \(\frac{13-2\sqrt{3}}{6}>\sqrt{2}\)
a, \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\Rightarrow1+1< \sqrt{2}+1\Rightarrow2< \sqrt{2}+1\)
c, \(4>3=>\sqrt{4}>\sqrt{3}=>\sqrt{4}-1>\sqrt{3}-1\Rightarrow1>\sqrt{3}-1\)
d, \(16>11=>\sqrt{16}>\sqrt{11}\Rightarrow4>\sqrt{11}=>4.\left(-3\right)< \sqrt{11}.\left(-3\right)\)
\(=>-12< -3.\sqrt{11}\)
Lời giải:
a.
$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$
$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.
$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$
$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$
$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$