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\(\dfrac{x^2-4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\dfrac{x+2}{x}\)
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\(=\dfrac{\left(x-2\right)\left(x+1\right)}{-\left(x-2\right)\left(x+2\right)}=\dfrac{-x-1}{x+2}\)
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\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\\ =x\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^{16}-y^{16}\right)+xy^{16}\\ =x^{17}-xy^{16}+xy^{16}\\ =x^{17}\)
\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\)
\(=x\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^{16}-y^{16}\right)+xy^{16}\)
\(=x^{17}-xy^{16}+xy^{16}\)
\(=x^{17}\)
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\(\dfrac{x^5+x^3+x^2+1}{x^3+x^2+x+1}=\dfrac{x^3\left(x^2+1\right)+\left(x^2+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)
= \(\dfrac{\left(x^3+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}=x^2-x+1\)
\(\dfrac{x^5+x^3+x^2+1}{x^3+x^2+x+1}=\dfrac{x^3.\left(x^2+1\right)+\left(x^2+1\right)}{x.\left(x^2+1\right)+\left(x^2+1\right)}\) \(=\dfrac{\left(x^3+1\right).\left(x^2+1\right)}{\left(x+1\right).\left(x^2+1\right)}=\dfrac{x^3+1}{x+1}=\dfrac{\left(x+1\right).\left(x^2-x+1\right)}{x+1}\) \(=x^2-x+1\)
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Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
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Câu 4:
\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)
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\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x-4}{x-2}\)