rut gon (x^2+2x+1)\(x-1)^2:(2x^2+4x+2)\(4x^2-8x+4)
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1) \(\dfrac{x^2-18x-19}{x^2-1}=\dfrac{x^2-19x+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-19\right)+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-19\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-19}{x-1}\)
2) \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}=\dfrac{4x\left(x^2-2x+1\right)}{2x^2\left(x-1\right)}=\dfrac{4x\left(x-1\right)^2}{2x^2\left(x-1\right)}=\dfrac{2\left(x-1\right)}{x}\)
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a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)
\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)
\(=-19x^2+16x-4\)
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Ta có 2x(2x + 1)2 - 3x(x + 3)(x - 3) - 4x(x + 1)2
= 2x(4x2 + 4x + 1) - 3x(x2 - 9) - 4x(x2 + 2x + 1)
= 8x3 + 8x2 + 2x - 3x3 + 27x - 4x3 - 8x2 - 4x
= 8x3 - 3x3 - 4x3 + 8x2 - 8x2 + 2x + 27x - 4x
= x3 + 25x
a oi hinh nhu sai r con +16x2 nua co a , anh tinh lai ho e duoc kh
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\(P\left(x\right)=5+x^3-2x+4x^3+3x^2-10=\left(x^3+4x^3\right)+3x^2+2x-\left(10-5\right)=5x^3+3x^2+2x-5\)
\(Q\left(x\right)=4-5x^3+2x^2-x^3+6x-11x^3-8x=-\left(5x^3+x^3+11x^3\right)+2x^2-\left(8x-6x\right)+4=-17x^3+2x^2-2x+4\)
\(P\left(x\right)-Q\left(x\right)=\left(5x^3+3x^2+2x-5\right)-\left(-17x^3+2x^2-2x+4\right)=5x^3+3x^2+2x-5+17x^3-2x^2+2x-4\)
\(=\left(5x^3+17x^3\right)+\left(3x^2-2x^2\right)+\left(2x+2x\right)-\left(5+4\right)=22x^3+x^2+4x-9\)
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\(C=\dfrac{\sqrt{\dfrac{4x^2+4x+1}{x}}}{\sqrt{x}\cdot\left|2x^2-x-1\right|}=\dfrac{\left|2x+1\right|}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}\cdot\left|\left(x-1\right)\left(2x+1\right)\right|}\)
\(=\dfrac{1}{x\left|x-1\right|}\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}:\frac{2\left(x+1\right)^2}{4\left(x-1\right)^2}=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}.\frac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=2\)