a. x(2x2 – 3) – x2(5x + 1) + x2

      = 2x3 – 3x – 5x3 – x2 + x2 = -3x – 3x3

b. 3x(x – 2) – 5x(1 – x) – 8(x2 – 3)

      = 3x2 – 6x – 5x + 5x2 – 8x2 + 24

      = - 11x + 24

c. 1/2 x2(6x – 3) – x( x2 + 1/2 (x + 4)

      = 3x3 - 3/2 x2 – x3 - 1/2 x + 1/2 x + 2

      = 2x3 - 3/2 x2 + 2

a, x(2x2-3)-x2(5x+1)x2

=2x3-3x-5x3- x2+x2=-3x-3x3

học tốt nhé!!

b, 3x(x-2)-5x(1-x)-8(x2-3)

=3x2-6x-5x+5x2-8x2+24

=-11x+24

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

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

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

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

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

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

= (x - y)² - 2(x - y)(x + y) + (x + y)²

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

`@` `\text {Ans}`

`\downarrow`

Gửi c!

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) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

3x(x – 2) – 5x(1 – x) – 8( x 2  – 3)

= 3x.x + 3x .( -2) – [5x.1 + 5x. (- x)] – [8 x 2  + 8.(- 3)]

= (3 x 2  – 6x) – (5x – 5 x 2 ) – (8 x 2  – 24)

= 3 x 2  – 6x – 5x + 5 x 2  – 8 x 2 + 24

= ( 3 x 2  +5 x 2  – 8 x 2 )- ( 6x + 5x) + 24

= - 11x + 24

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

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