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a)\(A=\frac{x+1}{x^2-2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\left(\frac{1}{x\left(x-1\right)}+\frac{1}{x-1}\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{x+1}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{1+x}\)
\(=\frac{x}{x-1}\)
b)Có:\(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=2\end{array}\right.\)
Thay x=2 vào A ta được:
\(A=\frac{2}{2-1}=2\)
a/ ĐKXĐ ....
A=\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
=\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
=\(\frac{1}{x}-\frac{1}{x-5}\)
=\(-\frac{5}{x^2-5x}\)
b/ \(x^3-x+2=0\Leftrightarrow\left(x+1\right)\left(\left(x-1\right)^2+1\right)=0\)
<=> x=-1, thay vào tính nốt
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 ??
1. A = -4 phần x+2
2. 2x^2 + x = 0 => x = 0 hoặc x = -1/2
Với x = 0 thì A = -2
Với x = -1/2 thì A = -8/3
3. A = 1/2 => -4 phần x + 2 = 1/2
<=> -8 = x + 2
<=> x = -10
4. A nguyên dương => A > 0
=> -4 phần x + 2 > 0
Do -4 < 0 nên -4 phần x + 2 > 0 khi x + 2 < 0
=> x < -2
a, \(Đkxđ:\hept{\begin{cases}x\ne1\\x\ne\pm3\end{cases}}\)
\(P=\left(1+\frac{1}{x-1}\right):\left(\frac{x^2-7}{x^2-4x+3}+\frac{1}{x-1}+\frac{1}{3-x}\right)\)
\(=\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right):\left(\frac{x^2-7}{\left(x-1\right)\left(x-3\right)}+\frac{1}{x-1}-\frac{1}{x-3}\right)\)
\(=\left(\frac{x-1+1}{x-1}\right):\left(\frac{x^2-7+x-3-\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\right)\)
\(=\frac{x}{x-1}:\frac{x^2-7+x-3-x+1}{\left(x-1\right)\left(x-3\right)}\)
\(=\frac{x}{x-1}.\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}\)
\(=\frac{x}{x-1}.\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x}{x+3}\)
b, \(|x+2|=5\)
\(\Rightarrow x+2=\hept{\begin{cases}5\Leftrightarrow x+2\ge0\Rightarrow x\ge-2\\-5\Leftrightarrow x+2< 0\Rightarrow x< -2\end{cases}}\)
Nếu \(x\ge-2\Rightarrow x+2=5\)
\(\Rightarrow x=3\)\(\left(ktmđkxđ\right)\)
Nếu \(x< -2\Rightarrow x+2=-5\)
\(\Rightarrow x=-7\)\(\left(tm\right)\)
Vậy \(x=-7\)
a, \(A=\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{x+2}\right)\left(\frac{2}{x}-1\right)\)
\(=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\left(\frac{2-x}{x}\right)\)
\(=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}=\frac{-4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=\frac{-4}{x+2}\)
b, Ta có : \(2x^2+x=0\Leftrightarrow x\left(2x+1\right)=0\Leftrightarrow x=0;-\frac{1}{2}\)
Thay x = 0 vào biểu thức A ta được : \(\frac{-4}{0+2}=\frac{-4}{2}=-2\)
Thay x = -1/2 vào biểu thức A ta được : \(\frac{-4}{-\frac{1}{2}+2}=\frac{-4}{\frac{3}{2}}=-\frac{2}{3}\)
c, Ta có : \(\frac{-4}{x+2}=\frac{1}{2}\Leftrightarrow-8=x+2\Leftrightarrow x=-10\)
d, Ta có : \(\frac{-4}{x+2}\)hay \(x+2\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x + 2 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | -1 | -3 | 0 | -4 | 2 | -6 |
a)\(A=\frac{x+1}{x^2+2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x+1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{1}{x+1}\cdot\frac{x\left(x+1\right)}{x+1}=\frac{x}{x+1}\)
b)Có: \(x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=-2\end{array}\right.\)
Thay x=-2 vào A ta có
\(A=\frac{-2}{-2+1}=\frac{-2}{-1}=2\)