Toán $D=\sqrt{9x^{2}-2x}(x<0)$ $x-4+\sqrt{16-8x+x^{2}}(x>4)$ 10/07/2021 By Caroline $D=\sqrt{9x^{2}-2x}(x<0)$ $x-4+\sqrt{16-8x+x^{2}}(x>4)$
D = $\sqrt{9x^{2} }$ = $\sqrt{(3x)^{2} }$ =$ |3x| $ = $ -3x$ (Vì x < 0) Ta có: $x – 4 $+ $\sqrt{16-8x+x^{2} }$ = x – 4 + $\sqrt{(x-4)^{2}}$ =$ x- 4 + |x – 4 |$ = $x – 4 + x – 4$ (vì x > 4) = $2x – 8$ Trả lời
`x-4+\sqrt{16-8x+x^2}(x>4)` `=x-4+\sqrt{(x-4)^2}` `=x-4+|x-4|` `=x-4+x-4=2x-8` `\sqrt{9x^2}(x<0)` `=|3x|(x<0)` `=-3x` Trả lời
D = $\sqrt{9x^{2} }$
= $\sqrt{(3x)^{2} }$
=$ |3x| $
= $ -3x$ (Vì x < 0)
Ta có:
$x – 4 $+ $\sqrt{16-8x+x^{2} }$
= x – 4 + $\sqrt{(x-4)^{2}}$
=$ x- 4 + |x – 4 |$
= $x – 4 + x – 4$ (vì x > 4)
= $2x – 8$
`x-4+\sqrt{16-8x+x^2}(x>4)`
`=x-4+\sqrt{(x-4)^2}`
`=x-4+|x-4|`
`=x-4+x-4=2x-8`
`\sqrt{9x^2}(x<0)`
`=|3x|(x<0)`
`=-3x`