K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1 tháng 12 2018

ĐK: \(\hept{\begin{cases}x^2+4x+4\ne0\\4-x^2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\ne0\\\left(2-x\right)\left(2+x\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}}\)

\(P=\frac{x^3-4x}{x^2+4}.\left(\frac{1}{x^2+4x+4}+\frac{1}{4-x^2}\right)\)

\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{1}{\left(x+2\right)^2}+\frac{-1}{\left(x-2\right)\left(x+2\right)}\right)\)

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

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

30 tháng 6 2017

a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)

=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)

\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)

\(=\frac{4x}{\left(x+1\right)^2}\)=VP

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

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

=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP

c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)

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

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

\(=x+y=\)VP

Vậy các đẳng thức được chứng minh

=

30 tháng 6 2017

C là xy mà ko phải x+y

11 tháng 1 2022

\(A=\left(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}+\dfrac{4}{4x+16}\right):\dfrac{1}{4x}\left(x\ne4;x\ne-4;x\ne0\right).\)

\(A=\left(\dfrac{1}{x\left(x-4\right)}+\dfrac{-2}{\left(x+4\right)\left(x-4\right)}+\dfrac{1}{x+4}\right).4x\).

\(A=\dfrac{x+4-2x+x^2-4x}{x\left(x-4\right)\left(x+4\right)}.4x.\)

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

\(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}.4.\)

\(A=\dfrac{4\left(x-1\right)}{x+4}.\)

 

11 tháng 1 2022

chịch ko em

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)