đề bài là j hả bn

ta có:

(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)

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

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

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

= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

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

= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)

= \(\dfrac{6-x}{x-6}\)

= \(\dfrac{-\left(x-6\right)}{x-6}\)

= -1

a) 4x(x - 5) - (x - 1)(4x - 3) = 5

4x² - 20x - (4x² - 3x - 4x + 3) = 5

4x² - 20x - 4x² + 3x + 4x - 3 = 5

-13x - 3 = 5

\(\Rightarrow\) -13x = 8

\(\Rightarrow\) x = \(\dfrac{-8}{13}\)

b) (3x - 4)(x - 2) = 3x(x - 9) - 3

3x² - 6x - 4x + 8 = 3x² - 27x - 3

3x² - 10x + 8 - 3x² + 27x + 3 = 0

17x + 11 = 0

\(\Rightarrow\) 17x = -11

\(\Rightarrow\) x = \(\dfrac{-11}{17}\)

c) x² - 81 = 0

\(\Rightarrow\) x² = 81

\(\Rightarrow\) x = \(\pm\) 9

d) 3x² - 75 = 0

3(x² - 25) = 0

\(\Rightarrow\) x² - 25 = 0

\(\Rightarrow\) x² = 25

\(\Rightarrow\) x = \(\pm\)5

e) x² - 4x + 3 = 0

x² - x - 3x + 3 = 0

(x² - x) - (3x - 3) = 0

x(x - 1) - 3(x - 1) = 0

(x - 3)(x - 1) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

xin lỗi vì chữa đề

chứng minh ba điểm đó tạo thành 1 góc bằng 180 độ

i) 7x² - 7xy - 5x + 5y

= (7x² - 7xy) - (5x - 5y)

= 7x(x - y) - 5(x - y)

= (7x - 5)(x - y)

bn xem lại dề ài nhé

Ta có:

\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)

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

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

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

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

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

= \(\dfrac{1}{x-y}\) (đpcm)

\(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)