Câu 2:
1.
ĐKXĐ: \(x\ne0;x\ne\pm2\)
\(R\left(\dfrac{x-1}{x^2-2x}+\dfrac{x+1}{x^2+2x}-\dfrac{4}{x^3-4x}\right):\dfrac{4026}{x}\\ =\left[\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x+2\right)\left(x-2\right)}\right]\cdot\dfrac{x}{4026}\\ =\left[\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right]\cdot\dfrac{x}{4026}\\ =\dfrac{x^2-x+2x-2+x^2-2x+x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x}{4026}\\ =\dfrac{2x^2-8}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x}{4026}\\ =\dfrac{2\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x}{4026}\\ =\dfrac{2}{4026}\\ =\dfrac{1}{2013}\)
Câu 1:
1: \(P=2a^3+7a^2b+7ab^2+2b^3\)
\(=2\left(a^3+b^3\right)+7ab\left(a+b\right)\)
\(=2\left(a+b\right)\left(a^2-ab+b^2\right)+7ab\left(a+b\right)\)
\(=\left(a+b\right)\left(2a^2-2ab+2b^2+7ab\right)\)
\(=\left(a+b\right)\left(2a^2+4ab+ab+2b^2\right)\)
\(=\left(a+b\right)\left[2a\left(a+2b\right)+b\left(a+2b\right)\right]\)
\(=\left(a+b\right)\left(a+2b\right)\left(2a+b\right)\)
2: \(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)
\(=x^6+x^5-x^4+x^5+x^4-x^3+2x^4+2x^3-2x^2+x^3+x^2-x+3x^2+3x+1\)
\(=x^4\left(x^2+x-1\right)+x^3\left(x^2+x-1\right)+2x^2\left(x^2+x-1\right)+x\left(x^2+x-1\right)+3\left(x^2+x\right)+1\)
=3+1=4
Câu 2:
2: TH1: x>=2
Phương trình sẽ trở thành:
(x-2)(x-1)(x+1)(x+2)=4
=>\(\left(x^2-4\right)\left(x^2-1\right)=4\)
=>\(x^4-5x^2+4-4=0\)
=>\(x^4-5x^2=0\)
=>\(x^2\left(x^2-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=\sqrt{5}\left(nhận\right)\\x=-\sqrt{5}\left(loại\right)\end{matrix}\right.\)
TH2: x<2
Phương trình sẽ trở thành:
(2-x)(x-1)(x+1)(x+2)=4
=>(x-2)(x-1)(x+1)(x+2)=-4
=>\(\left(x^2-4\right)\left(x^2-1\right)+4=0\)
=>\(x^4-5x^2+8=0\)
=>\(x^4-2\cdot x^2\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}=0\)
=>\(\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)
