MT
1
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
JS
1
9 tháng 5 2017
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3^2\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3+x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-\frac{3}{x+3}\right).\frac{x+3}{3x^2}\)
\(A=-x^2\)
HT
0
7 tháng 12 2014
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
B
2
TA
5 tháng 12 2021
ĐK: `x \ne 3; x \ne -3`
`A=3/(x-3)-(6x)/(9-x^2)+x/(x+3)`
`=3/(x-3)+(6x)/(x^2-9)+x/(x+3)`
`=3/(x-3)+(6x)/((x-3)(x+3))+x/(x+3)`
`=(3(x+3)+6x+x(x-3))/((x-3)(x+3))`
`=(3x+9+6x+x^2-3x)/((x+3)(x-3))`
`=(x^2+6x+9)/((x-3)(x+3))`
`=((x+3)^2)/((x-3)(x+3))`
`=(x+3)/(x-3)`
`x=5 => A=(5+3)/(5-3)=4`
5 tháng 12 2021
ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\9-x^2\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x^2\ne9\\x\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(3-x\right)\left(3+x\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x+3}{x-3}\)
Thay x=5 vào \(\dfrac{x+3}{x-3}=\dfrac{5+3}{5-3}=\dfrac{8}{2}=4\)
21 tháng 7 2019
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
QL
5
\(\Leftrightarrow\frac{3\left(x-3\right)+3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}.\frac{\left(x-3\right)^2}{9}\)
\(\Leftrightarrow\frac{3x-9+3x+9}{x+3}.\frac{x-3}{9}\)
\(\Leftrightarrow\frac{6x\left(x-3\right)}{9\left(x+3\right)}\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{3\left(x+3\right)}\)