Ta có: \(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\)

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

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

TA có: \(\left(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\right):\left(\frac{1}{x}-\frac{1}{y}\right)^2\)

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

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

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

a: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

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

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

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

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

b:Sửa đề: \(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\)

\(=\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\)

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

\(=\frac{2x-2-2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{-6}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

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

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

\(\dfrac{x^3-8}{x+7}\left(\dfrac{3}{x-2}-\dfrac{2}{x^2+2x+4}\right)\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\left(\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\right)\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\left(\dfrac{3x^2+6x+12}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x^2+2x+4\right)}\right)\)

\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}.\dfrac{3x^2+6x+12-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}\)

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

\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x+7}\cdot\dfrac{3x^2+6x+12-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}\)

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

mk chịu !!!!

ai làm đk giúp mik vs ạ

a. 50-17+2-50+15

= (50-50)+(15+2-17)

= 0+0 = 0

a.0

b.79
c.16400
d.10

\(\frac{3\left(x-2\right)}{4}\div\frac{2-x}{2}=\frac{3\left(x-2\right)}{4}\times\frac{-2}{x-2}=\frac{-3}{2}\)

học tốt

Rút gọn nhé !

\(\frac{3}{4}.\left(x-2\right):\frac{1}{2}.\left(2-x\right)=\frac{3x-6}{4}.2.\left(2-x\right)\)

\(=\frac{3x-6}{4}.\left(4-2x\right)=\frac{\left(3x-6\right).\left(4-2x\right)}{4}\)

\(=\frac{\left(12x-24\right)-\left(6x^2+12x\right)}{4}=\frac{-24-6x^2}{4}\)

\(=\frac{-12-3x^2}{2}=\frac{-3.\left(4+x^2\right)}{2}\)

giúp em  bài với ạ,em cảm ơn, em đang vội ạ

\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)

\(-23-150+\left(-9\right)+1913\)

\(=1913-\left(23+150+9\right)\)

\(=1913-182\)

\(=1731\)

a, 23 + ( -77 ) + ( -23 ) + 77

= [ 23 + ( -23 ) ] + [ ( -77 ) + 77 ]

= 0 + 0 = 0

b, ( -2020 ) + 2021 + 21 + ( -22 )

= ( - 2020 ) + 2021 + 21 + ( -21 ) + ( -1 )

= [ ( -2020 ) + 2021 + ( -1 ) ] + [ 21 + ( -21 ) ]

= 0 + 0 = 0

a) 23 + (-77) + (-23) + 77 =
[23 + (-23)] + [(-77) + 77]
=0+0=0
b) (-2 020) + 2 021 + 21 + (-22)
=[(-2 020) + 2 021] + [21 + (-22)]
=1+(-1)
= 0.