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\(\text{a, ĐKXĐ: }\hept{\begin{cases}x+3\ne0\\x-3\ne0\\3x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\mp3\\x\ne0\end{cases}}\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)
\(=\frac{x-x-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
b, với x=\(-\frac{1}{2}\)ta có:
\(A=-\frac{1}{\left(-\frac{1}{2}\right)^2}=-4\)
c, Để A<0 thì \(-\frac{1}{x^2}< 0\text{ mà }x^2>0\left(\text{vì x khác 0 ĐKXĐ}\right)\)
Với x khác 0 thì thỏa mãn!
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\frac{\left(3-x\right)\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=\frac{3\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
a) \(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\) (ĐKXĐ: \(x\ne\pm1\) )
\(=\left(\frac{x+1+2\left(1-x\right)-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\frac{2}{x^2-1}.\frac{x^2-1}{1-2x}=\frac{2}{1-2x}\)
b) Để x nhận giá trị nguyên <=> 2 chia hết cho 1 - 2x
<=> 1-2x thuộc Ư(2) = {1;2;-1;-2}
Nếu 1-2x = 1 thì 2x = 0 => x= 0
Nếu 1-2x = 2 thì 2x = -1 => x = -1/2
Nếu 1-2x = -1 thì 2x = 2 => x =1
Nếu 1-2x = -2 thì 2x = 3 => x = 3/2
Vậy ....
\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)
\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)
\(A=\frac{-1}{x^2}\)
Ta có :\(x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)
\(\Rightarrow A=\frac{-1}{2^2}\)
\(A=\frac{-1}{4}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
Rảnh hả bạn?? =))