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1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)
=\(\frac{-3+x}{\left(x-2\right)\left(x-3\right)}.\frac{x+1}{1-x}\)
=\(\frac{1}{\left(x-2\right)}.\frac{x+1}{1-x}\)
=\(\frac{x+1}{\left(x-2\right)\left(1-x\right)}\)
b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)
\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)
\(\Leftrightarrow\frac{x-3}{x-2}>0\)
\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)
\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)
Vậy ...
BÀI 1:
a) \(ĐKXĐ:\) \(x-3\)\(\ne\)\(0\)
\(\Leftrightarrow\)\(x\)\(\ne\)\(3\)
b) \(A=\frac{x^3-3x^2+4x-1}{x-3}\)
\(=\frac{\left(x^3-3x^2\right)+\left(4x-12\right)+11}{x-3}\)
\(=\frac{x^2\left(x-3\right)+4\left(x-3\right)+11}{x-3}\)
\(=x^2+4+\frac{11}{x-3}\)
Để \(A\)có giá trị nguyên thì \(\frac{11}{x-3}\)có giá trị nguyên
hay \(x-3\)\(\notinƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta lập bảng sau
\(x-3\) \(-11\) \(-1\) \(1\) \(11\)
\(x\) \(-8\) \(2\) \(4\) \(14\)
Vậy....
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Để C=0 thì x-1=0
hay x=1
c: Để C>0 thì x-1>0
hay x>1
Vậy: \(\left\{{}\begin{matrix}x\in Z\backslash\left\{1\right\}\\x\notin\left\{2;-2\right\}\end{matrix}\right.\)
c) Để A nhận giá trị nguyên khi và chỉ khi:
Kết hợp với điều kiện, tập hợp các giá trị của x nguyên để A nguyên là: {0; 2; -2; 4}.