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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$H=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-16)$
$=x^3-3x^2+3x-1-x^3-8+3x^2-48$
$=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-48)$
$=3x-57=3.\frac{-1}{2}-57=\frac{-117}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
`4*(2y+3x)-3(x-3y)`
`=8y+12x-3x+9y`
`=8y+9y+12x-3x`
`=17y+9x`
b)
`x^2 +2x-x(7x-3)`
`=x^2 +2x-7x^2 +3x`
`=x^2 -7x^2 +2x+6x`
`= -6x^2 +8x`
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 4.( x - 3) - 3|x + 3|
- Nếu x > -3 ta có A = 4.(x - 3) - 3.(x + 3) = 4x - 12 - 3x - 9 = x - 3
- Nếu x < -3 ta có A = 4.(x - 3) - 3.(-x - 3) = 4x - 12 + 3x + 9 = x + 21
B = 2.|x + 1| - |x - 1|
- Nếu x > 1 thì B = 2.(x + 1) - (x - 1) = 2x + 2 - x + 1 = x + 3
- Nếu x = 0 thì B = 2.(0 + 1) - (0 - 1) = 2 - (-1) = 3
- Nếu x < 0 thì B = 2.(-x - 1) - (-x + 1) = -2x - 2 + x - 1 = -x - 3
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(P=4\left(x-3\right)-3\left|x+3\right|\)
Trường hợp 1: x>=-3
\(P=4x-12-3x-9=x-21\)
Trường hợp 2: x<-3
P=4x-12+3x+9=7x-3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(|x|-x\)
\(\Rightarrow\orbr{\begin{cases}x< 0\rightarrow\left|x\right|-x=2\left|x\right|\\x>0\rightarrow\left|x\right|-x=0\end{cases}}\)
\(\Rightarrow x=0\rightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(B=\left|x-\dfrac{1}{7}\right|-\left|x+\dfrac{3}{5}\right|+\dfrac{4}{5}\)
\(=-x+\dfrac{1}{7}-x-\dfrac{3}{5}+\dfrac{4}{5}\)
\(=-2x+\dfrac{12}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ai giúp mình bài này với
Ta có : \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|4-x\right|\ge0\forall x\\\left|x\right|\ge0\forall x\end{cases}\Rightarrow}\hept{\begin{cases}\left|x+3\right|=x+3\\\left|4-x\right|=4-x\\\left|x\right|=x\end{cases}}\)
\(\Rightarrow3\left|x-3\right|+2\left|4-x\right|+\left|x\right|\)
\(=3.\left(x-3\right)+2.\left(4-x\right)+x\)
\(=3x-9+8-2x+x\)
\(=\left(3x-2x+x\right)-\left(9-8\right)\)
\(=2x+1\)