ko biết

\(2x^3\left(x^2-5\right)+\left(-2x^3+4x\right)+\left(6+x\right)x^2\)

\(=2x^5-10x^3-2x^3+4x+6x^2+x^3=2x^5-9x^3+6x^2+4x\)

ko biet

(-3).8/8.6 rút gọn

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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

Bài 4: 

5x(4x² - 2x + 1) - 2x(10x² - 5x - 2)

= 5x.4x² + 5x.(-2x) + 5x.1 - 2x.10x² + (-2x).(-5x) + (-2x).(-2)

= 20x³ - 10x² + 5x - 20x³ + 10x² + 4x

= (20x² - 20x²) + (-10x² + 10x²) + (5x + 4x)

= 0 + 0 + 9x

= 9x (1)

Thay x = -15 vào (1), ta có:

9.(-15) = -135

Vậy: Giá trị biểu thức sau khi rút gọn với x = -15 là: -135

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)

\(=\left(-3\right)^2+5\left(2x-3\right)\)

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

\(a)\)

\(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(4x-6\right)+xy\)

\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2+xy\)

\(=\left(2x+3-2x+3\right)^2+xy\)

\(=6^2+2\left(-1\right)\)

\(=36-2\)

\(=34\)

\(b)\)

\(\left(x-2\right)^2-\left(x-1\right)\left(x+1\right)-x\left(1-x\right)\)

\(=x^2-4x+4-x^2+1-x+x^2\)

\(=x^2-5x+5\)

Thay \(x=-2\)vào ta có:

\(\left(-2\right)^2-5\left(-2\right)+5\)

\(=4+10+5\)

\(=19\)

Bài 1:

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)