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

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

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

\(=\frac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\frac{10}{x}\)

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

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

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

\(=\frac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\frac{10}{x}\)

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

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

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

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

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

\(=\frac{10x^2+10y^2}{x\left(x^2+y^2\right)}\)

\(=\frac{10}{x}\)

đề bài thực hiện phép tính ( / ) là phân số ai nhanh mình k cố lên nhé

hình như sai đề rồi bn

Xin lỗi bài này sai

a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)

\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)

\(=\frac{3x^2-3y^2}{50}\)

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

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

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

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

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

d) \(\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)

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

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

\(=\frac{2x^2+2y^2}{x^2-y^2}.\frac{x^2+y^2}{y^2}\)

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

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

Tui học lớp 7 đăng bài lớp 8 không được sao?