cho biểu thức P=\(\left(\frac{x^4+x^2-4x+1}{x^2}-\frac{x-1}{x+1}+\frac{x+1}{x-1}\right).\left(\frac{x\left(x+1\right)-\left(1+x\right)}{x^3-1}\right)\)
a)tìm x để P xác định
b>rút gọn P
c>tìm giá trị x để P nguyên
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 6 2019
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.
21 tháng 4 2020
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1
DL
5 tháng 10 2019
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
