me ajudem a resolver a) e b) da minha lição
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a)
![(2x^2-3x+4)-[(5x^2+3x-5)-(x^2+1)]=\\
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2x^2-3x+4-[4x^2+3x-6]=\\
\\
\boxed{-2x^2-6x+10}](https://tex.z-dn.net/?f=%282x%5E2-3x%2B4%29-%5B%285x%5E2%2B3x-5%29-%28x%5E2%2B1%29%5D%3D%5C%5C%0A%5C%5C%0A2x%5E2-3x%2B4-%5B4x%5E2%2B3x-6%5D%3D%5C%5C%0A%5C%5C%0A%5Cboxed%7B-2x%5E2-6x%2B10%7D "(2x^2-3x+4)-[(5x^2+3x-5)-(x^2+1)]=\\
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2x^2-3x+4-[4x^2+3x-6]=\\
\\
\boxed{-2x^2-6x+10}")
b)
![(2y^2-3y+5)-[(5y^2+3y-7)-(y^2+3)]=\\
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2y^2-3y+5-[4y^2+3y-10]=\\
\\
\boxed{-2y^2-6y-5}](https://tex.z-dn.net/?f=%282y%5E2-3y%2B5%29-%5B%285y%5E2%2B3y-7%29-%28y%5E2%2B3%29%5D%3D%5C%5C%0A%5C%5C%0A2y%5E2-3y%2B5-%5B4y%5E2%2B3y-10%5D%3D%5C%5C%0A%5C%5C%0A%5Cboxed%7B-2y%5E2-6y-5%7D "(2y^2-3y+5)-[(5y^2+3y-7)-(y^2+3)]=\\
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2y^2-3y+5-[4y^2+3y-10]=\\
\\
\boxed{-2y^2-6y-5}")
b)
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