a) = (x+3).(x-3)^2-(x-3)(x+3)^2

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

=(x^2-9)(x-3-x-3)

=-6(x^2-9)

các câu còn lại tương tự

\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

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

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

\(=6\)

\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)

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

\(=-125-125\)

\(=-250\)

a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)

\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)

\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)

b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)

\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)

\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)

c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)

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

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

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

d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)

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

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

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

a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)

\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)

\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)

\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)

b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)

\(=4x^2-16.\dfrac{7x-2}{3x+6}\)

\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)

c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)

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

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

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

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

d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)

\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)

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

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

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

Giúp tui vs mn

lớp 8 có pt bậc 2 ak??

Có nhưng giải bằng PT tích nhé

Bạn gõ ct lại dưới dạng latex (kí hiệu này gọi là latex Σ ) nhé

a, 3y-2y=2y-3

    3y-2y-2y=3

    -y=3

     y=-3

b, 3-4x+24+6x=x+27+3x

   -4x+6x-x-3x =27-3-24

   -2x              =0

      x             =0

c, 5-(6-x)=4.(3-2x)

   5-6+x =12-8x

   x+8x  =12+6-5

  9x      =13

   x       =13/9

d, 4.(x+3)=-7x+17

   4x+12  =-7x+17

4x+7x     =17-12

11x         =5

  x          =5/11

a: \(\Leftrightarrow\left(3x+2\right)\left(5-x\right)=-9x^2+4\)

\(\Leftrightarrow\left(3x+2\right)\left(5-x\right)+\left(3x+2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(2x+3\right)=0\)

=>x=-2/3 hoặc x=-3/2

b: \(\Leftrightarrow4x\left(x+5\right)+x^2-25=0\)

\(\Leftrightarrow\left(x+5\right)\left(5x-5\right)=0\)

=>x=-5 hoặc x=1

c: \(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

=>x=1 hoặc x=-1/2