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

b: \(=-5y-9+xy\)

\(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\)

b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\)

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

\(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\)

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

\(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\)

a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)

Yêu cầu đề là gì vậy bạn?

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

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

\(=x^3+3x^2y+3xy^2+y^3\)

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

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

\(=x^3-3x^2y+3xy^2-y^3\)

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

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

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

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

\(a,=12x^3y-4x^2y^2+3xy^3\\ b,=x^3+3x^2-5x+3x^2+9x-15-x^3-4x^2+4x\\ =2x^2+8x-15\)

b: Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)-x\left(x-2\right)^2\)

\(=x^3+3x^2-5x+3x^2+9x-15-x^3+4x^2-4x\)

\(=10x^2-15\)

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

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

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

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].

phần e là cả hai dòng nhé các bạn

bn viết rõ đề đi bn

Vd:x2 là 2.x hay x\(^2\)

Có nhiều chỗ vậy lắm bn ạ,bn viết lại đề đi rồi tụi mk giúp cho.

a) \(3x-3y+x^2-y^2\)

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

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

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

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

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

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

c) \(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)

↓

\(=\left(x^2-y^2-2y-1\right)\left(x^2-2xy+y^2-9\right)\)

\(=\left[x^2-\left(y^2+2y+1\right)\right]\left(x^2-2xy+y^2-9\right)\)

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

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

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

d) \(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

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

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

\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)

\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)

e)

- \(9x^2+90=9\left(x+10\right)\)

- \(x+225-\left(x-7\right)^2\)

\(=x+225-\left(x^2-14x+49\right)\)

\(=x+225-x^2+14x-49\)

\(=-x^2+15x+176\)

\(=-\left(x^2-15x-176\right)\)

a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)

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

\(=5x^3-7x^2y+2xy^2+5x-2y\)

b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

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

c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)

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