Matemática, perguntado por inteligencia111, 9 meses atrás

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Respondido por 1gatinha1
0

a)p\left(x\right)\:\mathrm{de}\:\frac{p\left(x\right)eA}{\left(x-1\right)^2}+\left(p\left(x\right)-2\right)^2=5:\\\p\left(x\right)=\frac{-8x-eA+4x^2+4+\sqrt{20x^4-80x^3+120x^2-8eAx^2+16eAx-80x+e^2A^2+20-8eA}}{2\left(x^2+1-2x\right)}

\frac{p\left(x\right)eA}{\left(x-1\right)^2}+\left(p\left(x\right)-2\right)^2=5\\\\\frac{p\left(x\right)eA}{\left(x-1\right)^2}\left(x-1\right)^2+\left(p\left(x\right)-2\right)^2\left(x-1\right)^2=5\left(x-1\right)^2;\quad \:x\ne \:1\\\\\\eAp\left(x\right)+\left(p\left(x\right)-2\right)^2\left(x-1\right)^2=5\left(x-1\right)^2;\quad \:x\ne \:1\\\\eAp\left(x\right)+\left(p\left(x\right)-2\right)^2\left(x-1\right)^2:\quad \\\\

\mathrm{Expandir\:}eAp\left(x\right)+\left(p\left(x\right)-2\right)^2\left(x-1\right)^2:\quad eAp\left(x\right)+x^2p\left(x\right)^2-2xp\left(x\right)^2+p\left(x\right)^2-4x^2p\left(x\right)+\\\\eAp\left(x\right)+x^2p\left(x\right)^2-2xp\left(x\right)^2+p\left(x\right)^2-4x^2p\left(x\right)+8xp\left(x\right)-4p\left(x\right)+4x^2-8x+4=5x^2-10x+5;\quad \:x\ne \:1

eAp\left(x\right)+x^2p\left(x\right)^2-2xp\left(x\right)^2+p\left(x\right)^2-4x^2p\left(x\right)+8xp\left(x\right)-4p\left(x\right)+4x^2-8x+4-\left(5x^2-10x+5\right)=5x^2\\\\\left(x^2+1-2x\right)p\left(x\right)^2+\left(8x+eA-4x^2-4\right)p\left(x\right)+2x-x^2-1=0;\quad \:x\ne \:1

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=x^2+1-2x,\:b=8x+eA-4x^2-4,\:c=2x-x^2-1:\quad p\left(x\right)_{1,\:2}=\frac{-\left(8x+eA-4x^2-4\right)\pm \sqrt{\left(8x+eA-4x^2-4\right)^2-4\left(x^2+1-2x\right)\left(2x-x^2-1\right)}}{2\left(x^2+1-2x\right)}

p\left(x\right)=\frac{-8x-eA+4x^2+4+\sqrt{20x^4-80x^3+120x^2-8eAx^2+16eAx-80x+e^2A^2+20-8eA}}{2\left(x^2+1-2x\right)}\\\left(x\right)=\frac{-8x-eA+4x^2+4+\sqrt{20x^4-80x^3+120x^2-8eAx^2+16eAx-80x+e^2A^2+20-8eA}}{2\left(x^2+1-2x\right)};\:x\ne \:1


1gatinha1: eu fiz só a letra A tbm
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