a. x +3 + √x^2 – 6x +9 (x ≤ 3) b. √x^2 + 4x +4 – √x^2 (-2 ≤ x ≤ 0) 06/09/2021 Bởi Allison a. x +3 + √x^2 – 6x +9 (x ≤ 3) b. √x^2 + 4x +4 – √x^2 (-2 ≤ x ≤ 0)
a. x +3 + $\sqrt[]{x^2 – 6x +9}$ (x ≤ 3) = x + 3 + $\sqrt[]{(x-3)²}$ = x + 3 + x – 3 = 2x b. √(x^2 + 4x +4) – √x^2 (-2 ≤ x ≤ 0) = √( x +2)² – √x^2 = x+2 – x = 2 Bình luận
`a. x +3 + sqrt(x^2 – 6x +9 )(x ≤ 3)` `=x+3+sqrt((x-3)^2)` `=x+3+x-3` `=2x` `b. sqrt(x^2 + 4x +4 )- sqrt(x^2 )(-2 ≤ x ≤ 0)` `=sqrt((x+2)^2)-x` `=x+2-x` `=2` Bình luận
a. x +3 + $\sqrt[]{x^2 – 6x +9}$ (x ≤ 3)
= x + 3 + $\sqrt[]{(x-3)²}$
= x + 3 + x – 3
= 2x
b. √(x^2 + 4x +4) – √x^2 (-2 ≤ x ≤ 0)
= √( x +2)² – √x^2
= x+2 – x
= 2
`a. x +3 + sqrt(x^2 – 6x +9 )(x ≤ 3)`
`=x+3+sqrt((x-3)^2)`
`=x+3+x-3`
`=2x`
`b. sqrt(x^2 + 4x +4 )- sqrt(x^2 )(-2 ≤ x ≤ 0)`
`=sqrt((x+2)^2)-x`
`=x+2-x`
`=2`