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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 \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
Lời giải:
ĐKXĐ: \(x\neq \left\{2;\pm 3\right\}\)
a) Ta có:
\(P=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x(x-3)}{(x-3)(x+3)}-1\right):\left(\frac{(3-x)(3+x)}{(x-2)(x+3)}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x}{x+3}-1\right):\left(\frac{3-x}{x-2}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\frac{x-(x+3)}{x+3}:\left(-\frac{x-2}{x+3}\right)=\frac{-3}{x+3}.\frac{x+3}{-(x-2)}=\frac{3}{x-2}\)
b) \(x^3-3x+2=0\)
\(\Leftrightarrow (x^3-x)-2(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-2(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-2)=0\)
\(\Leftrightarrow (x-1)[(x^2-1)+(x-1)]=0\)
\(\Leftrightarrow (x-1)^2(x+2)=0\) \(\Leftrightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x=1\Rightarrow P=\frac{3}{1-2}=-3\)
Với \(x=-2\Rightarrow P=\frac{3}{-2-2}=\frac{-3}{4}\)
c)
\(P=\frac{3}{x-2}\in\mathbb{Z}\Leftrightarrow 3\vdots x-2\)
\(\Leftrightarrow x-2\in \text{Ư}(3)\Rightarrow x-2\in\left\{\pm 1; \pm 3\right\}\)
\(\Leftrightarrow x\in \left\{3,1,5,-1\right\}\)
Do \(x\neq 3\Rightarrow x\in \left\{-1,1,5\right\}\)
a: ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)
b: \(A=\left(\dfrac{x}{x-3}-\dfrac{2x-1}{x\left(x-3\right)}\right)\cdot\dfrac{x-3}{1}\)
\(=\dfrac{x^2-2x+1}{\left(x-3\right)\cdot x}\cdot\dfrac{x-3}{1}=\dfrac{\left(x-1\right)^2}{x}\)
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
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) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\dfrac{x^2-3x+9}{x-3}.\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=-\dfrac{3}{x-3}\)
Đề sai ạ ! Sửa lại nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right)\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-x^2+3x-9}\)
\(\Leftrightarrow A=\frac{-\left(x+3\right)}{x}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow-\left(x+3\right)⋮x\)
\(\Leftrightarrow-x-3⋮x\)
\(\Leftrightarrow3⋮x\)
\(\Leftrightarrow x\inƯ\left(3\right)\)
Vậy để \(A\inℤ\Leftrightarrow x\inƯ\left(3\right)\)(\(x\neℤ\))
Bạn sửa cho mik dòng cuối :
\(x\ne Z\)thành \(x\notin Z\)nhé !
a) giải phương trình
\(\dfrac{2x^2-3x-2^{ }}{_{ }x^2-4}\) = 2
=>\(\dfrac{2x^2-3x-2}{x^2-4}\) = \(\dfrac{2\left(x^2-4\right)}{x^2-4}\)
=>2x2 - 3x - 2 = 2(x2 - 4)
<=>2x2 -3x - 2 = 2x2 - 8
<=>2x2 - 2x2 - 3x = -8 + 2
<=>-3x = -6
<=> x = 2
Vậy không tồn tại giá trị nào của x thỏa mãn điều kiện của bài toán
b) Ta phải giải phương trình
\(\dfrac{6x-1}{3x+2}\) = \(\dfrac{2x+5}{x-3}\)
=>x = \(\dfrac{-7}{38}\)
c) Ta phải giải phương trình
\(\dfrac{y+5}{y-1}\) - \(\dfrac{y+1}{y-3}\) = \(\dfrac{-8}{\left(y-1\right)\left(y+1\right)}\)
không tồn tại giá trị nào của y thỏa mãn điều kiện của bài toán
\(ĐKXĐ:x\ne0;x\ne\pm3\)
\(A=\left(\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right):\left(\frac{x^2}{3\left(9-x^2\right)}+\frac{1}{x+3}\right)\\ =\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{-3\left(x+3\right)\left(x-3\right)}{x^2-3x+9}\\ =\frac{-x-3}{x}\)
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
\(A=\left(\dfrac{1}{3x+2}+\dfrac{1}{3x-2}\right):\dfrac{1}{3x+2}\)
\(A=\left(\dfrac{1\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}\right).3x+2\) \(A=\dfrac{3x-2+3x+2}{\left(3x-2\right)\left(3x+2\right)}.3x+2\)
\(A=\dfrac{6x.\left(3x+2\right)}{\left(3x+2\right)\left(3x-2\right)}\)
\(A=\dfrac{6x}{3x-2}\)