\(\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}\)

\(=\dfrac{3\left(1-x\right)}{\left(x+1\right)^2}:\dfrac{6\left(x^2-1\right)}{x+1}\)

\(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}:\dfrac{6\left(x+1\right)\left(x-1\right)}{x+1}\)

\(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{-3\left(x-1\right)\left(x+1\right)}{6\left(x+1\right)^3\left(x-1\right)}=\dfrac{-3\left(x+1\right)}{6\left(x+1\right)\left(x+1\right)^2}=\dfrac{-3}{6\left(x+1\right)^2}=\dfrac{-1}{2\left(x+1\right)^2}\)

b) Bạn có thể viết kiểu latex được không ạ ?

cảm ơn xong chẳng có ai :)))

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

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

\(=x^2-x\)

2: \(2x\left(3x+2\right)-3x\left(2x+3\right)\)

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

=-5x

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

\(=x^4-18x^2+81-x^4+81=-18x^2+162\)

b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)

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

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

1) \(-6x^4+4x^3-2x^2\)

2) \(=x^2+4x-21-x^2-4x+5=-16\)

3) \(=6x^2-4x-x^2-4x-4=5x^2-8x-4\)

4) \(=2x^3-4x^2-8x-3x^2+6x+12=2x^3-7x^2-2x+12\)

giúp mình với

làm như bình thường là đc:

  a)   ( x + 2 )^3 - x^2  . ( x+ 6 ) - 8

= ( x^3 + 3.x^2.2 +3.x.2^2 + 2^3 ) - ( x^3 + 6x^2 ) - 8

= x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 - 8

=12x

b)     ( x - 2 ) . ( x^2 + 2x + 4 ) - ( x^3 + 2 )

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

= x^3 - 8 - x^3 - 2

= -6

Bạn học đến hằng đẳng thức chưa

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].