KT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
0
NL
1
ND
1 tháng 6 2018
\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)\(\left(ĐKXĐ:x\ne\pm\dfrac{1}{3}\right)\)
\(=\left[\dfrac{-3x\left(3x+1\right)+2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-3x^2-5x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-x\left(3x+5\right).\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right).2x\left(3x+5\right)}\)
\(=\dfrac{1-3x}{2\left(3x+1\right)}\)
\(=\dfrac{1-3x}{6x+2}\)
29 tháng 7 2021
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
DD
2
27 tháng 7 2018
\(ĐKXĐ:1-3x\ne0\Leftrightarrow x\ne\dfrac{1}{3};3x+1\ne0\Leftrightarrow x\ne-\dfrac{1}{3}\)
\(A=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x}\)
\(A=\left(\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\right):\left(\dfrac{6x^2+10x}{1+3x}\right)\)
\(A=\dfrac{3x^2+5x}{3x+1-9x^2-3x}.\dfrac{1+3x}{6x^2+10}\)
\(A=\dfrac{\left(3x^2+5x\right).\left(1+3x\right)}{\left(1-9x^2\right).2.\left(3x^2+5\right)}\)
\(A=\dfrac{1+3x}{\left(1+3x\right)\left(1-3x\right)}=\dfrac{1}{\left(1-3x\right).2}=\dfrac{1}{2-6x}\)
27 tháng 7 2018
Hình như đề bị thiếu mũ \(2\) trong \(1-6x+9x\)\(\) đúng không Đinh Diệp
\(A=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(A=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1+3x\right)\left(1-3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1+3x\right)\left(1-3x\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x+9x^2}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x-6x^2}{\left(1+3x\right)\left(1-3x\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\left[\dfrac{3x^2+5x}{-\left(3x-1\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\left[\dfrac{x\left(3x+5\right)}{-\left(3x-1\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\dfrac{1}{-\left(1+3x\right)}.\dfrac{3x-1}{x}\)
\(A=\dfrac{1}{-1-3x}.\dfrac{3x-1}{x}\)
\(A=\dfrac{3x-1}{-x-3x^2}\)
LA
26 tháng 2 2019
a) Điều kiện : \(x\ne\pm\dfrac{1}{3}\)
\(B=\left[\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right]:\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\left(\dfrac{3x\left(3x+1\right)}{\left(1-3x\right)\left(3x+1\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right):\dfrac{6x^2+10x}{ \left(3x-1\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\cdot\dfrac{\left(1-3x\right)^2}{6x^2+10x}\)
\(=\dfrac{x\left(3x+5\right)}{\left(1-3x\right)\left(3x+1\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}=\dfrac{1-3x}{2\left(3x+1\right)}\)
b) Sai đề = Không làm
c) B >0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x>0\\2\left(3x+1\right)>0\end{matrix}\right.\\\left[{}\begin{matrix}1-3x< 0\\2\left(3x+1\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x>-\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
TH1 => \(-\dfrac{1}{3}< x< \dfrac{1}{3}\)
TH2 :Vô lí
Vậy giá trị x thỏa mãn :
\(-\dfrac{1}{3}< x< \dfrac{1}{3}\)
ĐKXĐ: \(x\ne\left\{-\dfrac{1}{3};\dfrac{1}{3};0;-\dfrac{4}{3}\right\}\)
\(M=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\left(\dfrac{x\left(3x+5\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{1-3x}{2\left(1+3x\right)}\)