Đáp án:
`a)` `\sqrt{2}+\sqrt{3}<\sqrt{10}`
`b)` `2+\sqrt{3}<\sqrt{2}+\sqrt{6}`
`c)` `16>\sqrt{15}.\sqrt{17}`
Giải thích các bước giải:
`a)` Ta có:
`\qquad (\sqrt{2}+\sqrt{3})^2`
`=2+2\sqrt{6}+3=5+2\sqrt{6}<5+2.\sqrt{{25}/4}`
`<5+2. 5/2=10`
`=>(\sqrt{2}+\sqrt{3})^2<10`
`=>\sqrt{2}+\sqrt{3}<\sqrt{10}`
$\\$
`b)` `(2+\sqrt{3})^2`
`=4+2.2\sqrt{3}+3=7+4\sqrt{3}`
`\qquad (\sqrt{2}+\sqrt{6})^2`
`=2+2.\sqrt{12}+6=8+2.2\sqrt{3}`
`=8+4\sqrt{3}>7+4\sqrt{3}`
`=>(2+\sqrt{3})^2<(\sqrt{2}+\sqrt{6})^2`
`=>2+\sqrt{3}<\sqrt{2}+\sqrt{6}`
$\\$
`c)` `\sqrt{15}.\sqrt{17}`
`=\sqrt{15.17}`
`=\sqrt{(16-1).(16+1)}`
`=\sqrt{16^2-1^2}<\sqrt{16^2}=16`
Vậy `16>\sqrt{15}.\sqrt{17}`
