Câu này  cô làm rồi em nhá, em xem phần câu hỏi của tôi ý

Q = \(\dfrac{1+x^4+x^8+...+x^{2020}}{1+x^2+...+x^{2022}}\)

Đặt A = 1 + \(x^4\) + \(x^8\) +...+ \(x^{2020}\)

Đặt B = 1 + \(x^2\) + ...+ \(x^{2022}\)

Thì Q = \(\dfrac{A}{B}\) 

A              = 1 + \(x^4\) + \(x^8\) + ...+ \(x^{2020}\)

A.\(x^4\)         =       \(x^4\) + \(x^8\) +....+ \(x^{2020}\) + \(x^{2024}\)

A.\(x^4\) - A    = \(x^{2024}\) - 1

A              = \(\dfrac{x^{2024}-1}{x^4-1}\) 

B             = 1 + \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\) 

B.\(x^2\)        =       \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\) + \(x^{2024}\)

B\(x^2\) - B   =       \(x^{2024}\) - 1

B             = \(\dfrac{x^{2024}-1}{x^2-1}\)

Q = \(\dfrac{\dfrac{x^{2024}-1}{x^4-1}}{\dfrac{x^{2024}-1}{x^2-1}}\)

Q  = \(\dfrac{x^{2024}-1}{x^4-1}\) \(\times\)\(\dfrac{x^2-1}{x^{2024}-1}\)

Q  = \(\dfrac{1}{x^2+1}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+...+\dfrac{1}{x+2020}-\dfrac{1}{x+2022}\)

\(=\dfrac{x+2022-x}{x\left(x+2022\right)}=\dfrac{2022}{x\left(x+2022\right)}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(=\dfrac{x-4}{x-2}\)

Làm chi tiết theo yêu cầu của bạn :

\(ĐKXĐ\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(=\dfrac{x-4}{x-2}\)

p/s : bonus nếu bạn không hiểu ptich đa thức thành nhân tử =))

x² + x - 6 = x² + 3x - 2x - 6 = x(x+3) - 2(x+3) = (x-2)(x+3)

x² - x - 12 = x² + 3x - 4x - 12 = x(x+3) - 4(x+3) = (x-4)(x+3)

\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

\(A=\dfrac{x-1}{x+1}\)

c) Thay x = 3 vào A ta có:

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

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

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

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

B=4/9(1/3+2/3)+14/9=4/9+14/9=2/1

B = 2/1 = 2

B=4/9(1/3+2/3)+14/9

B=4/9+14/9

B=2/1

\(=\dfrac{1-x}{xy\left(x-1\right)}=\dfrac{-1}{xy}\)

Câu 4: 

\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)