Alternante:i, a! (2x3+5xi)-3x2=4x5
ax4-15=(80-75)x5
(6x4-ix3)+9x4=3x10+3x3
96-(8x4+ax3...
Mathematics, 06.01.2021 15:40 mprjug6
Alternante:i, a! (2x3+5xi)-3x2=4x5
ax4-15=(80-75)x5
(6x4-ix3)+9x4=3x10+3x3
96-(8x4+ax3)=5x6+19
96-(9x5-9xi)=5x6+57
96-(5x5+ax6-40)x2=9x10
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Answers: 3
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