Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
=5x^2+5x-2x-2-(5x^2+x-15x-3)-17x-51
=5x^2-14x-53-5x^2+14x+3
=-50
Lời giải:
$H=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-16)$
$=x^3-3x^2+3x-1-x^3-8+3x^2-48$
$=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-48)$
$=3x-57=3.\frac{-1}{2}-57=\frac{-117}{2}$
+ 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
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right).x^2.\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
Ta có:
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)x^2\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left[\left(2x^3-x^2+1\right)+\left(x^2-2x^3\right)\right]\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
A = (x - 2)2 + (x + 3)2 - 2.(x + 1)(x - 1)
A = (x - 2)2 + (x + 3)2 - 2.(x2 - 1)
A = x2 - 2.2x + 22 + x2 + 2.3x + 32 - 2x2 + 2
A = 2x + 15
\(A=\left(x-2\right)^2+\left(x+3\right)^2-2\left(x+1\right)\left(x-1\right)\)
\(A=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(A=2x^2+2x+13-2x^2+2\)
\(A=2x+15\)
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
x<-1 nên x+1<0
x<-1 nên x<2
=>x-2<0
\(N=\left|2-x\right|-3\left|x+1\right|\)
\(=\left|x-2\right|-3\left|x+1\right|\)
\(=2-x-3\left(-x-1\right)\)
=2-x+3x+3=2x+5