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2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
a, \(A=\left(\frac{x}{x+3}+\frac{x}{x-3}-\frac{2}{x^2-9}\right).\frac{x+3}{2x-2}\)
\(=\frac{x^2-3x+x^2+3x-2}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2\left(x-1\right)}=\frac{2\left(x-1\right)\left(x+1\right)\left(x+3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+3\right)}=\frac{x+1}{x-3}\)
Ta có : A = 2 hay \(\frac{x+1}{x-3}=2\Rightarrow x+1=2x-6\Leftrightarrow-x=-7\Leftrightarrow x=7\)(tmđk )
b, \(A< 0\Rightarrow\frac{x+1}{x-3}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>3\end{cases}}}\)( vô lí )
TH2 : \(\hept{\begin{cases}x+1>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow-1< x< 3}}\)
Kết hợp với đk ta được -1 < x < 3 ; x khác 1
a ) ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm3\end{matrix}\right.\)
\(A=\left(\frac{x^2-3}{x^2-9}+\frac{x+3}{x^2-9}\right)\cdot\frac{x+3}{x}\)
\(A=\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x}\)
\(A=\frac{x\left(x+1\right)}{x-3}\cdot\frac{1}{x}=\frac{x+1}{x-3}\)
b) \(\left|A\right|=3\Leftrightarrow\left[{}\begin{matrix}A=3\\A=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-3}=3\\\frac{x+1}{x-3}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+1=3x-9\\x+1=-3x+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\4x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\) ( TM ĐKXĐ )
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
a) \(ĐKXĐ:x\ne\pm3\)
b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)
\(\Leftrightarrow A=\frac{-3}{x+3}\)
c) Tại \(x=-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)
\(\Leftrightarrow A=\frac{-6}{5}\)
d) Để \(A>0\)
\(\Leftrightarrow\frac{-3}{x+3}>0\)
\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)
\(\Leftrightarrow x< -3\)
e) +) Với \(A>\frac{-1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)
\(\Leftrightarrow-6>-x-3\)
\(\Leftrightarrow x>3\)(tm)
+) Với \(A< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)
\(\Leftrightarrow-6< -x-3\)
\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))
+) Với \(A=-\frac{1}{2}\)
\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)
\(\Leftrightarrow x+3=6\)
\(\Leftrightarrow x=3\)(ktm)
Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)