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ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x^2+x+1\ne0\end{cases}}\)
a/ \(R=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right]\)
\(=1:\left[\frac{x^2+2+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x}{x^2+x+1}\right)\)
\(=\frac{x^2+x+1}{x}\)
b/ Ta có: \(R=\frac{x^2+x+1}{x}=3+\frac{\left(x-1\right)^2}{x}>3\)
Vậy R > 3
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
Nãy ấn nhầm thông cảm
1) a) đkxđ \(x\ne\pm3,x\ne1\)
Ta có : \(P=\left(\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^2+3}{x^2-9}\right):\left(\frac{2x-2}{x-3}-1\right)\)
\(=\left(\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+3}{\left(x+3\right)\left(x-3\right)}\right):\frac{2x-2-x+3}{x-3}\)
\(=\frac{2x^2-6x+x^2+3x-3x^2-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+1}{x-3}\)
\(=\frac{-3x-3}{\left(x+3\right)\left(x-3\right)}.\frac{x-3}{x+1}=\frac{-3\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}=\frac{-3}{x+3}\)
b) Để \(P\in Z\) thì \(\frac{-3}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(-3\right)=\left\{\pm1,\pm3\right\}\)
Ta có bảng giá trị
| x+3 | 1 | -1 | 3 | -3 |
| x | -2 | -4 | 0 |
-6 |
Vậy với \(x\in\left\{-2,-4,0,6\right\}\) thì \(P\in Z\)
c) \(\left|x+3\right|=5\Leftrightarrow\left[{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Thay x=2 vào P, ta có : \(P=-\frac{3}{2+2}=-\frac{3}{4}\)
Thay x=-8 vào P, ta có : \(P=-\frac{3}{-8+2}=\frac{1}{2}\)
Vậy ....
2) a) đkxđ : \(x\ne1\)
Ta có : \(R=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=1:\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=1:\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+x+1}{x}\)
Xét : \(P-3=\frac{x^2+x+1}{x}-3=\frac{x^2-2x+1}{x}=\frac{\left(x-1\right)^2}{x}\)
+)Nếu \(x\ge0,x\ne1\Rightarrow R>3\)
+) Nếu \(x< 0\Rightarrow R< 3\)
+) Nếu \(\left[{}\begin{matrix}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{matrix}\right.\) \(\Rightarrow R=3\)
c) Để \(R>4\Rightarrow\frac{x^2+x+1}{x}>4\) \(\Rightarrow x^2+x+1>4x\)
\(\Rightarrow x^2>3x-1\) \(\Rightarrow x>\frac{3x-1}{x}=3-\frac{1}{x}\)
Vậy \(x>3-\frac{1}{x}thìR>4\)
d) Thay x=1/4 vào R, ta có : \(R=\frac{\frac{1}{16}+\frac{1}{4}+1}{\frac{1}{4}}=\frac{21}{4}\)
đề bài mk cảm thấy nó sao sao í bạn ạ hoặc do mk tính sai
a: R-3=(x^2+x-1-3x)/x=(x-1)^2/x
Nếu x>0 thì R-3>0
=>R>3
Nếu x<0 thì R-3<0
=>R<3
c: Để R>4 thì R-4>0
=>\(\dfrac{x^2+x+1-4x}{x}>0\)
=>\(\dfrac{x^2-3x+1}{x}>0\)
TH1: x>0 và x^2-3x+1>0
=>x>0 và \(\left[{}\begin{matrix}x< \dfrac{3-\sqrt{5}}{2}\\x>\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow x>\dfrac{3+\sqrt{5}}{2}\)
mà x nguyên
nên x>3
TH2: x<0 và x^2-3x+1<0
=>x<0 và \(\dfrac{3-\sqrt{5}}{2}< x< \dfrac{3+\sqrt{5}}{2}\)(loại)