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2 tháng 8 2016

a) 4.(x + 2) = 2 .(x - 1) + 4

=> 4x + 8 = 2x - 2 + 4

=> 4x + 8 = 2x + 2

=> 4x - 2x = 2 - 8

=> 2x = -6

=> x = -3

b) (x + 1) + ( 2x + 2) + (3x + 3) + ( 4x + 4) = 80

=> (x + 2x + 3x + 4x ) + (1 + 2 + 3 + 4) = 80

=> 10x + 10 = 80

=> 10x = 70

=> x = 7

2 tháng 8 2016

a, 4.(x+2) = 2.(x-1) +4

  4x +8 = 2x-2+4

 4x-2x =-2+4-8

      2x = -6

        x = -3

b, (x+1)+(2x+2)+ ( 3x+3) + ( 4x+4) =80

 (x+1)+ 2 .(x+1) +3. (x+1) +4. (x+1) =80

(1+2+3+4) (x+1)=80

10.(x+1) =80

  x+1 =8 

  x=7

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

Bài 2: 

Ta có: \(P=3x\left(\dfrac{2}{3}x^2-3x^4\right)+9x^2\left(x^3-1\right)+x^2\left(-2x+9\right)-12\)

\(=2x^3-9x^5+9x^5-9x^2-2x^3+9x^2-12\)

=-12

Bài 1: 

a: Ta có: \(x\left(x^2+2\right)+2x\left(1-\dfrac{1}{2}x^2\right)=4\)

\(\Leftrightarrow x^3+2x+2x-x^3=4\)

hay x=1

b: Ta có: \(4x^2\left(x-1\right)+x\left(x^2+4x\right)=40\)

\(\Leftrightarrow4x^3-4x^2+x^3+4x^2=40\)

\(\Leftrightarrow5x^3=40\)

hay x=2

c: Ta có: \(3x\left(x-2\right)-3\left(x^2-3\right)=8\)

\(\Leftrightarrow3x^2-6x-3x^2+9=8\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)

3 tháng 8 2020

Bài 1:

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

\(=12x^2-4x-6x-2-x-3\)

\(=12x^2-11x-5\)

b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)

\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)

\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)

\(=2x^4y+1x^3y^2-2x^2y^3\)

c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)

\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)

\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)

\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)

\(=-48x^3-8x^2-24x^4+15x-9\)

3 tháng 8 2020

Bài 2 ạ

a: \(=12x^2-9x-\left(12x^2+10x-6x-5\right)\)

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

=-13x+5

b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)

\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)

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

c: \(=x^3-3x^2+3x-1-\left(x^3-8\right)+3\left(x^2-16\right)\)

\(=x^3-3x^2+3x-1-x^3+8+3x^2-48\)

\(=3x-41\)

6 tháng 8 2019

1a) -3x2(2x3 - 2x + 1/3) = -6x5 + 6x3 - x2

b) (x4 + 2x3 - 2/3).(-3x4) = -3x8 - 6x7 + 2x4

c) (x + 3)(x - 4) = x2 - 4x + 3x - 12 = x2 - x - 12

d)(x - 4)(x2 + 4x + 16) = (x - 4)(x2 + 4x + 42) = x3 - 64

e) 4(x - 1/2)(x + 1/2)(4x2 + 1) =4(x2 - 1/4)(4x2  + 1) = 4(4x4 + x2 - x2 - 1/4) = 4(4x4 - 1/4) = 16x4 - 1

B2. a) (2 - x)(x2 + 2x + 4) + x(x - 3)(x + 4) - x2 + 24 = 0

=> 8 - x3 + x(x2 + 4x - 3x - 12) - x2 + 24 = 0

=> 8 - x3 + x3 + x2 - 12x - x2 + 24 = 0

=> -12x + 32 = 0

=> -12x = -32

=> x = -32 : (-12) = 8/3

b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0

=> 5x/2 - 3x2 + 15 - 18x + 3x2 + 36x - x/2 - 6 = 0

=> 20x + 9 = 0

=> 20x = -9

=> x = -9/20