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\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
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\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1
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Bài 2 :
Câu a : \(10x^3y^3-15x^2y^4+20x^3=5x^2\left(2xy^3-3y^4+4x\right)\)
Câu b : \(9x^3-18x^2y+9xy^2=9x\left(x^2-2xy+y^2\right)=9x\left(x-y\right)^2\)
Câu c : \(9x^2+6x+1-25y^2=\left(3x+1\right)^2-\left(5y\right)^2=\left(3x+5y+1\right)\left(3x-5y+1\right)\)
Câu d : \(x^2-3x-18=\left(x^2+3x\right)-\left(6x+18\right)=x\left(x+3\right)-6\left(x+3\right)=\left(x+3\right)\left(x-6\right)\)