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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)
Ta có: P=|3x-6|-3x+5
=3x-6-3x+5
=-1
Ta có: Q=|8-2x|+3x+8
=2x-8+3x+8
=5x
![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
=5x^2+5x-2x-2-(5x^2+x-15x-3)-17x-51
=5x^2-14x-53-5x^2+14x+3
=-50
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|4-x\right|\ge0\forall x\\\left|x\right|\ge0\forall x\end{cases}\Rightarrow}\hept{\begin{cases}\left|x+3\right|=x+3\\\left|4-x\right|=4-x\\\left|x\right|=x\end{cases}}\)
\(\Rightarrow3\left|x-3\right|+2\left|4-x\right|+\left|x\right|\)
\(=3.\left(x-3\right)+2.\left(4-x\right)+x\)
\(=3x-9+8-2x+x\)
\(=\left(3x-2x+x\right)-\left(9-8\right)\)
\(=2x+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = 0 - 10
<=> x = -10
=> x = -10
Bài 3:
6(3q + 4q) - 8(5p - q) + (p - q)
= 6.3p + 6.4q - 8.5p - (-8).q + p - q
= 18p + 24q - 40p + 8q + p - q
= (18p - 40p + p) + (24q + 8q - q)
= -21p + 31q
![](https://rs.olm.vn/images/avt/0.png?1311)
+ Nếu x < -3, ta có:
3(x - 1) - 2|x + 3| = 3(x - 1) - 2(-3 - x)
= 3x - 3 + 6 + 2x
= 5x + 3
+ Nếu \(x\ge-3\) ta có:
3(x - 1) - 2|x + 3| = 3(x - 1) - 2(x + 3)
= 3x - 3 - 2x - 6
= x - 9
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)