Trong mặt phẳng Oxy cho ba đường tròn :

\(\left(C_1\right):\left(x-1\right)^2+\left(y-3\right)^2=4\)

\(\left(C_2\right):\left(x+3\right)^2+\left(y-4\right)^2=4\)

\(\left(C_3\right):\left(x+1\right)^2+\left(y-5\right)^2=5\)

Trong hai đường tròn \(\left(C_2\right)\) và \(\left(C_3\right)\), đường tròn là ảnh của \(\left(C_1\right)\) qua phép tịnh tiến. Xác định phép tịnh tiến này ?

Mina ơi~~~Ai giải giùm em vài bài này với a~~Em làm rùi nhưng cứ thấy hoang mang quá nên hỏi mina cho chắc a~Em cảm ơn mina nhiều a~

Bài 1:

b,\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

c,\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

d,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)

c,\(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

d,\(\dfrac{xy}{ab}+\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}-\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)

e,\(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)

g,\(\left(\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}\right).\left(\dfrac{x^2+y^2}{2xy}+1\right).\dfrac{xy}{x^2+y^2}\)

h,\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

Cho hàm số \(y=\dfrac{1}{2x^2+x-1}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?

A. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2019}}{\left(2x-1\right)^{2020}}\right)\)

B. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)

C. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2}{\left(2x-1\right)^{2020}}\right)\)

D. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{2}{\left(2x-1\right)^{2020}}\right)\)

Rút gọn các biểu thức sau : 

A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)

B = \(2x:\dfrac{1}{2}x+x^2\)

C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)

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

E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)

F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)

Rút gọn các biểu thức sau : 

A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)

B = \(2x:\dfrac{1}{2}x+x^2\)

C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)

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

E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)

F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)

Cho hàm số \(y=\dfrac{1}{3x^2-x-2}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?

A. \(\dfrac{2019!}{5}\left(\dfrac{1}{\left(x-1\right)^{2020}}-\dfrac{3}{\left(3x+2\right)^{2020}}\right)\)

B. \(\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

C. \(\dfrac{2019!}{5}\left(\dfrac{3}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

D. \(\dfrac{2019!}{5}\left(\dfrac{1}{\left(x-1\right)^{2020}}-\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}\right)\)