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a) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)
\(P=\frac{x}{x+1}\)
b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)
Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)
Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó:
\(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)
c) P > 1 khi \(\frac{x}{x+1}>1\)
\(\Leftrightarrow1-\frac{1}{x+1}>1\)
\(\Leftrightarrow\frac{1}{x+1}< 0\)
\(\Leftrightarrow x< -1\)
e) Đề không rõ ràng
a) \(p=\left(\frac{x^2-x}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}=2\)
b)\(m=\frac{x+2-\left(x-2\right)+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2}{x-2}=1+\frac{4}{x-2}\)
Để m nguyên thì \(4⋮x-2\)
\(\Rightarrow x-2\in\left\{1,2,4,-1,-2,-4\right\}\)
\(\Leftrightarrow x\in\left\{3,4,6,1,0,-2\right\}\)
\(M=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4x}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow M=\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow M=\frac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow M=\frac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{x-2}\)
Để M có giá trị nguyên thì x+2 chia hết cho x-2
Ta có x+2=x-2+4
=> x-2+4 chia hết cho x-2
=>4 chia hết cho x-2
Vì x nguyên => x-2 nguyên
=> x-2 thuộc Ư (4)={-4;-2;-1;1;2;4}
Ta có bảng
| x-2 | -4 | -2 | -1 | 1 | 2 | 4 |
| x | -2 | 0 | 1 | 3 | 4 | 6 |
a
\(ĐKXĐ:x\in R\)
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4-x^2+1\right)\)
\(=\frac{\left(x^2-1\right)\left(x^4-x^2+1\right)}{x^4-x^2+1}-\frac{x^4-x^2+1}{x^2+1}\)
\(=x^2-1-\frac{x^4-x^2+1}{x^2+1}\)
\(=-1+\frac{x^4+x^2-x^4+x^2+1}{x^2+1}\)
\(=\frac{2x^2+1}{x^2+1}-1=\frac{2x^2+1-x^2-1}{x^2+1}=\frac{x^2}{x^2+1}\)
b
Xét \(x>0\Rightarrow M>0\)
Xét \(x=0\Rightarrow M=0\)
Xét \(x< 0\Rightarrow M>0\)
Vậy \(M_{min}=0\) tại \(x=0\)
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)
\(=1-\frac{2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:
\(|x-\frac{3}{4}|=\frac{5}{4}\)
*TH1:
\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)
*TH2:
\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)
\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
c,
\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)
Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)
a, \(M=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(x-1\right)\left(x+1\right)=2x+1\)
b, Thay x = 1/2 vào biểu thức trên ta được : \(2.\frac{1}{2}+1=1\)
c, Để M luôn dương hay \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
Vậy với x \(\ge-\frac{1}{2}\)thì \(M\ge0\)