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a/ ĐKXĐ : \(x\ne0,3,1\)
\(P=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\dfrac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}.\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}.\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}.\dfrac{x}{2\left(x-1\right)}=-\dfrac{3}{x-1}\)
Vậy....
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
a)
ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)
Ta có: \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{x\left(x-3\right)}\right):\left(\dfrac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{x\left(x-3\right)}{3x\left(x-3\right)}+\dfrac{9}{3x\left(x-3\right)}\right):\left(\dfrac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{3x\left(x-3\right)}:\dfrac{-x^2+3x-9}{3\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{3x\left(x-3\right)}\cdot\dfrac{3\left(x-3\right)\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-x-3}{x}\)
b) Để A nguyên thì \(-x-3⋮x\)
mà \(-x⋮x\)
nên \(-3⋮x\)
\(\Leftrightarrow x\inƯ\left(-3\right)\)
\(\Leftrightarrow x\in\left\{1;-1;3;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-1\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{1;-1\right\}\)
a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b: \(P=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)
1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x+2\right)\left(x-2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)
\(\left(x+2\right)\ne0\Rightarrow x\ne-2\)
\(\left(x-2\right)\ne0\Rightarrow x\ne2\)
Vậy để biểu thức xác định thì : \(x\ne\pm2\)
b) để C=0 thì ....
1, c , bn Nguyễn Hữu Triết chưa lm xong
ta có : \(/x-5/=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)
thay x = 7 vào biểu thứcC
\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...
thay x = 3 vào C
\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)
=> ko tìm đc giá trị C tại x = 3