已知a² - 5a + 2 = 0.求2a4 – 10a³+ 7a²- 15a + 2020的值.

问题描述:

已知a² - 5a + 2 = 0.求2a4 – 10a³+ 7a²- 15a + 2020的值.

原式
a²-5a=-2
2a4-10a3+7a²-15a+2020
=2a²(a²-5a)+7a²-15a+2020
=2a²*-2+7a²-15a+2020
=3a²-15a+2020
=3(a²-5a)+2020
=3*-2+2020
=-6+2020
=2014

2a4 – 10a³+ 7a²- 15a + 2020
=2a的4次方-10a³+4a²+3a²-15a+6+2014
=2a²(a² - 5a + 2)+3(a² - 5a + 2)+2014
∵a² - 5a + 2=0
∴原式=2a²×0+3×0+2014=2014

a² - 5a + 2 = 0
a²-5a=-2
2a4 – 10a³+ 7a²- 15a + 2020
=2a²(a²-5a)+7a²-15a+2020
=-4a²+7a²-15a+2020
=3a²-15a+2020
=3(a²-5a)+2020
=-6+2020
=2014