Если x1 и x2-корни уравнения
2x^2+3x-4=0, то значение выражения x1^4+x2^4 равно ?
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Ответил Матов
0
По теореме Виета
![2x^2+3x-4=0\
x_{1}+x_{2}=-frac{3}{2}\
x_{1}x_{2}=-2\\
(x_{1}+x_{2})^4=x_{1}^4+4x_{1}x_{2}^3+6x_{1}^2x_{2}^2+4x_{1}^3x_{2}+x_{2}^4=frac{81}{16}\
4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=\
x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}=frac{9}{4}+2*2=frac{25}{4}\
4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=4*-2(frac{25}{4}) +6*4 = -50+24 = -26\
\
x_{1}^4+x_{2}^4=frac{81}{16}+26=frac{497}{16}](https://tex.z-dn.net/?f=2x%5E2%2B3x-4%3D0%5C%0Ax_%7B1%7D%2Bx_%7B2%7D%3D-frac%7B3%7D%7B2%7D%5C%0Ax_%7B1%7Dx_%7B2%7D%3D-2%5C%5C%0A%28x_%7B1%7D%2Bx_%7B2%7D%29%5E4%3Dx_%7B1%7D%5E4%2B4x_%7B1%7Dx_%7B2%7D%5E3%2B6x_%7B1%7D%5E2x_%7B2%7D%5E2%2B4x_%7B1%7D%5E3x_%7B2%7D%2Bx_%7B2%7D%5E4%3Dfrac%7B81%7D%7B16%7D%5C%0A4x_%7B1%7Dx_%7B2%7D%28x_%7B2%7D%5E2%2Bx_%7B1%7D%5E2%29%2B6x_%7B1%7D%5E2x_%7B2%7D%5E2%3D%5C%0Ax_%7B1%7D%5E2%2Bx_%7B2%7D%5E2%3D%28x_%7B1%7D%2Bx_%7B2%7D%29%5E2-2x_%7B1%7Dx_%7B2%7D%3Dfrac%7B9%7D%7B4%7D%2B2%2A2%3Dfrac%7B25%7D%7B4%7D%5C%0A4x_%7B1%7Dx_%7B2%7D%28x_%7B2%7D%5E2%2Bx_%7B1%7D%5E2%29%2B6x_%7B1%7D%5E2x_%7B2%7D%5E2%3D4%2A-2%28frac%7B25%7D%7B4%7D%29++%2B6%2A4++%3D+-50%2B24+%3D+-26%5C%0A%5C%0Ax_%7B1%7D%5E4%2Bx_%7B2%7D%5E4%3Dfrac%7B81%7D%7B16%7D%2B26%3Dfrac%7B497%7D%7B16%7D "2x^2+3x-4=0\
x_{1}+x_{2}=-frac{3}{2}\
x_{1}x_{2}=-2\\
(x_{1}+x_{2})^4=x_{1}^4+4x_{1}x_{2}^3+6x_{1}^2x_{2}^2+4x_{1}^3x_{2}+x_{2}^4=frac{81}{16}\
4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=\
x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}=frac{9}{4}+2*2=frac{25}{4}\
4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=4*-2(frac{25}{4}) +6*4 = -50+24 = -26\
\
x_{1}^4+x_{2}^4=frac{81}{16}+26=frac{497}{16}")
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