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Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
Cho biểu thức
A= (\( {1 \over x-2}\)+\({1 \over x+2}\)) : \( {5-x \over x-2}\)
a) Tìm ĐKXĐ
b) Rút gọn A
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đk : \(x\ne2;-3\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{x^2+x-6}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{x^2+x-6}\)
\(=\frac{x^2-x-12}{x^2+x-6}\)
\(=\frac{x^2-4x+3x-12}{x^2+3x-2x-6}\)
\(=\frac{x\left(x-4\right)+3\left(x-4\right)}{x\left(x+3\right)-2\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b)
A>0.
\(\frac{x-4}{x-2}>0\)
th1 :
x-4>0 và x-2>0
<=> x>4
th2 : x-4 <0 và x-2 < 0
<=> x<2
Vậy để A>0 thì x>4 hoặc x<2
a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\) \(\left(ĐKXĐ:x\ne2;-3\right)\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{-1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x^2-4x\right)+\left(3x-12\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x-4}{x-2}\)
b) Để \(A>0\)thì \(\frac{x-4}{x-2}>0\)
\(\Rightarrow\)(x - 4) ; (x - 2) cùng dấu
* hoặc \(\hept{\begin{cases}x-4>0\\x-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x>2\end{cases}}\Leftrightarrow x>4\)
* hoặc \(\hept{\begin{cases}x-4< 0\\x-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x< 2\end{cases}}\Leftrightarrow x< 2\)
Vậy \(\orbr{\begin{cases}x>4\\x< 2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ta có : \(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)\(=\frac{x+2}{x+3}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{1}{2-x}\)
\(=\frac{\left(x+2\right)\left(2-x\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{1\left(x+3\right)}{\left(x+3\right)\left(2-x\right)}\)
\(=\frac{-x^2+4+5+x+3}{\left(x+3\right)\left(2-x\right)}\) \(=\frac{-x^2+x+12}{\left(x+3\right)\left(2-x\right)}\)
\(=\frac{\left(4-x\right)\left(x+3\right)}{\left(x+3\right)\left(2-x\right)}\) = \(\frac{4-x}{2-x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(\frac{1-x^3+1-x-x}{1-x}\right):\frac{-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}=\left(\frac{-x^3-2x+2}{1-x}\right)\cdot\left(1-x\right)=-x^3-2x+2\)
b) \(-\left(-1\frac{2}{3}\right)-2\cdot\left(-1\frac{2}{3}\right)+2=\frac{5}{3}+\frac{10}{3}+2=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)