Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
a, Để A xác định
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne0\\x^2-9\ne\\3-x\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\left(x-3\right)\left(x+3\right)\ne0\\x\ne3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3;x\ne-3\\x\ne3\end{matrix}\right.\)
Vậy để A xác định \(\Leftrightarrow x\ne0;x\ne\pm3\)
b, Ta có: A \(=\dfrac{x-3}{2x}.\left(\dfrac{3x-1}{x^2-9}-\dfrac{1}{3-x}\right)\)
\(=\dfrac{x-3}{2x}\left[\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]\)
\(=\dfrac{x-3}{2x}.\dfrac{3x-1+x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x+2}{2x\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{2x\left(x+3\right)}=\dfrac{2x+1}{x\left(x+3\right)}\)
c) A=\(\dfrac{2x+1}{x\left(x+3\right)}=\dfrac{1}{2}\)
=>x(x+3)=2(2x+1)
=>x2+3x=4x+2
=>x2+3x-4x=2
=>x2-x-2=0
=> x2+x-2x-2=0
=>(x2+x)-(2x+2)=0
=>x(x+1)-2(x+1)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy A=\(\dfrac{1}{2}\) thì x=-1 hoặc x=2