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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: \(x\in\left\{-5;3;-3\right\}\)
\(A=\dfrac{-3\left(x+5\right)}{\left(x+5\right)^2}:\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+5}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-3\left(x+3\right)}\)
\(=\dfrac{x-3}{x+5}\)
b: Để A<1 thì A-1<0
=>\(\dfrac{x-3-x-5}{x+5}< 0\)
=>x+5>0
=>x>-5
c: Để A=(2x-3)/(x+1) thì \(\dfrac{2x-3}{x+1}=\dfrac{x-3}{x+5}\)
=>2x^2+10x-3x-15=x^2-2x-3
=>2x^2+7x-15-x^2+2x+3=0
=>x^2+9x-12=0
hay \(x=\dfrac{-9\pm\sqrt{129}}{2}\)
a) \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
\(=-\dfrac{9x^2+3x+2x-6x^2}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=-\dfrac{x\left(3x+5\right)}{\left(3x-1\right)^2}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-1}{2}\)
b) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\dfrac{x^2-3x+9}{x-3}.\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=-\dfrac{3}{x-3}\)
a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)
đkxđ: x khác 3, x khác -3
(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)
=>3x+9 -6x + x2+3x
<=>x2 + 3x-6x+3x + 9
<=>x2 +9
<=>(x-3).(x+3)
a) A \(=\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(\)\(=\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(=\dfrac{-3}{x+3}:\dfrac{3x^2}{x+3}\)
\(=\dfrac{-1}{x^2}\)
b) \(x=\dfrac{-1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne3;x\ne-3\) )
Thay \(x=\dfrac{-1}{2}\) vào biểu thức A, ta có:
\(A=\dfrac{-1}{\left(\dfrac{-1}{2}\right)^2}=-4\)
Vậy với \(x=\dfrac{-1}{2}\) giá trị của biểu thức A = -4.
c) \(\dfrac{-1}{x^2}< 0\)
\(\Rightarrow x^2>0\) (Luôn đúng)
Vậy với mọi giá trị của \(x\) để A < 0
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
\(M=\left(\dfrac{y}{x\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3\left(x-3\right)}{3.x\left(x+3\right)}-\dfrac{x^2}{3.x.\left(x+3\right)}\right)\\ =\left(\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3.x.\left(x+3\right)}\right)\\ =\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3.x.\left(x+3\right)}{3x-9-x^2}\\ =\dfrac{\left(y+x^2-3x\right).3}{\left(x-3\right)\left(3x-9-x^2\right)}\\ \)
a, ĐKXĐ: \(x\ne0;x\ne\pm3\)
Ta có: \(\left(\dfrac{y}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left[\dfrac{y}{x\left(x^2-9\right)}+\dfrac{1}{x+3}\right]:\left[\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right]\)
\(=\left[\dfrac{y}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right]:\left[\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right]\)
\(=\dfrac{y+1.x.\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{\left(x-3\right)\left(x+3\right)-x.x}{3x\left(x+3\right)}\)
\(=\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{\left(x-3\right)\left(x+3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{x^2-9-x^2}\)
\(=\dfrac{y+x^2-3x}{-3\left(x-3\right)}\)
\(ĐKXĐ:x\ne0;x\ne\pm3\)
\(A=\left(\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right):\left(\frac{x^2}{3\left(9-x^2\right)}+\frac{1}{x+3}\right)\\ =\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{-3\left(x+3\right)\left(x-3\right)}{x^2-3x+9}\\ =\frac{-x-3}{x}\)
b) Ta có :
\(A=\frac{-x-3}{x}< 1\\ \Leftrightarrow\frac{-x-3}{x}-1< 0\\ \Leftrightarrow\frac{-x}{x}-\frac{3}{x}-1< 0\\ \Leftrightarrow-1-1-\frac{3}{x}< 0\\ \Leftrightarrow-2-\frac{3}{x}< 0\\ \Leftrightarrow\frac{-3}{x}< 2\\ \Leftrightarrow2x< -3\\ \Rightarrow x>\frac{-3}{2}=-1,5\)
Vậy để A < 1 thì x > 1,5 / x ≠ 0 ; x ≠ 3 ; x ≠ -3