\(=\frac{15\left(x-y\right)^5}{5\left(x-y\right)^3}-\frac{10\left(x-y\right)^4}{5\left(x-y\right)^3}+\frac{20\left(x-y\right)^3}{5\left(x-y\right)^3}\)

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

sửa dòng cuối

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

a) =(x-y)⁵+(x-y)³=(x-y)³[(x-y)²+1]

b) =3³(y-2x)³:-9(y-2x)=-3(y-2x)²

c) =(x-y)² [3(x-y)³-2(x-y)²+3]:5(x-y)²=[3(x-y)³-2(x-y)²+3]/5

c) \(=x^2-4x+3-8x+2x^2-4+x-3x^2+2x-5\)

\(=-9x-6\)

b) \(=y^3-1+\frac{2}{3}x^3y-2xy+\frac{1}{3}x^2y^3-y^3\)

\(=\frac{2}{3}x^3y+\frac{1}{3}x^2y^3-2xy-1\)

a) (x² – 2x + 3) (x – 5)

= x³ - 5x² - x² +10x + x – 15

= x³ – 6x² + x -15

b) (x² – 2xy + y²)(x – y)

= x³ - x² y - 2x² y + 2xy² +xy²- y³

= x³ - 3x² y + 3xy² - y³

a) (x² – 2x + 3) ( 1/2x – 5) = \(\dfrac{1}{2}\)x³ – 5x² – x² + 10x +\(\dfrac{3}{2}\)x - 15

= \(\dfrac{1}{2}\)x³ – 6x² + \(\dfrac{23}{2}\) x – 15.

b) (x² – 2xy + y²)( x – y) = x³ – x²y – 2x²y + xy² – y³ = x³ – 3x²y + 3xy² – y³

a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)

=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)

=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)

=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)

=\(\frac{3x^3-4y}{24x^4y^5}\)

b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)

=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y-5x}{x\left(y+5x\right)}\)

c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)

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

=\(\frac{-2}{x\left(x-1\right)}\)

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

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

\(=\left(2y\right)^2=4y^2\)

Hằng đẳng thức thứ 2 nhé

ĐK: \(x,y\ne0,x\ne\pm y\)

Phép tính trên bằng:

        \(\left(\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{1}{x+y}.\frac{x^3-y^3}{xy}\right):\frac{x-y}{x}\)

\(=\left(\frac{\left(x-y\right)\left(x+y\right)^2}{xy\left(x+y\right)}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)xy}\right):\frac{x-y}{x}\)

\(=\left(\frac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}\right):\frac{x-y}{x}\)

\(=\frac{\left(x-y\right)xy}{xy\left(x+y\right)}.\frac{x}{x-y}=\frac{x}{x+y}\)

a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)

(tự rút gọn cái :P)

b, \(8x^3+4x^2y-2xy^2-y^3\)

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

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

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

Mấy cái còn lại nhân tung ra là được mà :))))

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