Trước hết, ta cần biết: `\sqrt{x^2}=?` Nếu `x>0` `⇒x^2>0` `⇒\sqrt{x^2}>0` `⇒\sqrt{x^2}=x` Nếu `x<0` `⇒x^2>0` `⇒\sqrt{x^2}>0` `⇒\sqrt{x^2}=-x` Nếu `x=0` `⇒\sqrt{x^2}=0` `⇒\sqrt{x^2}=x` `a,` Vì: `2>\sqrt3` `⇒2-\sqrt3>0` `⇒2\sqrt3+\sqrt{(2-\sqrt3)^2}` `=2\sqrt3+(2-\sqrt3)` `=(2\sqrt3-\sqrt3)+2` `=\sqrt3+2` `b, 2\sqrt{(-5)^6}+3\sqrt{(-2)^8}` `=2\sqrt{[(-5)^3]^2}+3\sqrt{[(-2)^4]^2}` `=2\sqrt{(-125)^2}+3\sqrt{16^2}` `=2.125+3.16` `=298` Trả lời
Trước hết, ta cần biết: `\sqrt{x^2}=?`
Nếu `x>0`
`⇒x^2>0`
`⇒\sqrt{x^2}>0`
`⇒\sqrt{x^2}=x`
Nếu `x<0`
`⇒x^2>0`
`⇒\sqrt{x^2}>0`
`⇒\sqrt{x^2}=-x`
Nếu `x=0`
`⇒\sqrt{x^2}=0`
`⇒\sqrt{x^2}=x`
`a,` Vì: `2>\sqrt3`
`⇒2-\sqrt3>0`
`⇒2\sqrt3+\sqrt{(2-\sqrt3)^2}`
`=2\sqrt3+(2-\sqrt3)`
`=(2\sqrt3-\sqrt3)+2`
`=\sqrt3+2`
`b, 2\sqrt{(-5)^6}+3\sqrt{(-2)^8}`
`=2\sqrt{[(-5)^3]^2}+3\sqrt{[(-2)^4]^2}`
`=2\sqrt{(-125)^2}+3\sqrt{16^2}`
`=2.125+3.16`
`=298`