bạn ơi , cái thứ 2 : x+ 27 làm j chung cái j đâu bạn
\(\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}\):\(\left(\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(\dfrac{x}{x^2+9}\):\(\left(\dfrac{x^2+9}{MTC}-\dfrac{6x}{MTC}\right)\)
\(\dfrac{x}{x^2+3^2}\):\(\dfrac{x^2-3x-3x+9}{MTC}\)
\(\dfrac{x}{x^2+3^2}\): \(\dfrac{x\left(x-3\right)-3\left(x-3\right)}{MTC}\)
\(\dfrac{x}{x^2+3^2}\): \(\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
\(\dfrac{x}{x^2+3^2}\). \(\dfrac{x^2+3^2}{x-3}\)
\(\dfrac{x}{x-3}\)
xong câu a rồi , đúng ko bạn
b)
\(\dfrac{x}{x-3}\)>= 0
x-3 > = 0
x>=3
chắc vậy nhỉ
làm bừa
a) \(\dfrac{x^2+3x}{\left(x^3+3x^2\right)+\left(9x+27\right)}\): \(\left(\dfrac{1}{x-3}-\dfrac{6x}{x^3-3x^2+x-27}\right)\)
\(\dfrac{x^2}{x^2\left(x+3\right)+9\left(x+3\right)}\): \(\left(\dfrac{1}{x-3}-\dfrac{6x}{x^2-\left(x-3\right)+x-27}\right)\)
đến đây tịt ngấm ngầm , bài gì mà hóc búa vậy
