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a.
\(ĐKXĐ:x\ne\pm1;\)
Ta có:
\(P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x-1}{x+1}+\frac{x+1}{x-1}\right)\cdot\frac{x\left(x+1\right)-\left(1+x\right)}{x^3-1}\)
\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)\left(x-1\right)}{x^3-1}\)
\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}+\frac{x^2+2x+1}{x^2-1}\right)\cdot\frac{x^2-1}{x^3-1}\)
\(\Rightarrow P=\frac{x^4+x^2+1}{x^2-1}\cdot\frac{x^2-1}{x^3-1}\)
\(\Rightarrow P=\frac{x^4+x^2+1}{x^3-1}\)
b.
Để P là số nguyên thì \(x^4+x^2+1⋮x^3-1\)
\(\Rightarrow\left(x^4-x\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x\left(x^3-1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x^2-x+1⋮x-1\)
\(\Rightarrow x\left(x-1\right)+1⋮x-1\)
\(\Rightarrow1⋮x-1\)
\(\Rightarrow x-1\in\left\{1;-1\right\}\)
\(\Rightarrow x=1\left(KTMĐK\right);x=0\)
Vậy x=0.
P/S:Không chắc chắn lắm đâu nha mn,nếu có j sai thì ib vs em ah.
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)
\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)
......................
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) \(p=\left(\frac{x^2-x}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}=2\)
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 |