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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\dfrac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+2\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\) (đpcm)
Dấu "=" xảy ra khi x = 1
Bn kia giải bài 1 r nên mk giải bài 2 nha!
Sửa lại:\(\dfrac{x^7+x^2+1}{x^8+x+1}\)
\(\dfrac{x^7+x^2+1}{x^8+x+1}=\dfrac{x^7-x+x^2+x+1}{x^8-x^2+x^2+x+1}\)
\(=\dfrac{x\left(x^6-1\right)+x^2+x+1}{x^2\left(x^6-1\right)+x^2+x+1}\)
\(=\dfrac{x\left(x^3-1\right)\left(x^3+1\right)+x^2+x+1}{x^2\left(x^3-1\right)\left(x^3+1\right)+x^2+x+1}\)
\(=\dfrac{x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x^2+x+1}{x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x^2+x+1}\)
\(=\dfrac{\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)}{\left(x^2+x+1\right)(x^6-x^5+x^3-x^2+1)}\)
Cả tử và mẫu đều có nhân tử:\(x^2+x+1>1\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự