已知3a的平方减a减一等于零,求6a的立方加7a的平方减5a加1999的值

3a²-a-1=0得3a²-a=1

6a³+7a²-5a+1999
=2a(3a²-a)+9a²-5a+1999
=2a+3(3a²-a)-2a+1999
=3+1999
=2002

3a²-a-1=0
3a²=a+1
3a²-a=1

6a的立方加7a的平方减5a加1999
=2a*a²+7a²-5a+1999
=2a(a+1)+7a²-5a+1999
=9a²-3a+199
=3(3a²-a)+199
=3+199
=202

因为3a^2-a-1=0，所以3a^2-a=1，
所以6a^3+7a^2-5a+1999=(6a^3-2a^2)+(9a^2-3a)-2a+1999=2a(3a^2-a)+3(3a^2-a)-2a+1999
=2a+3-2a+1999=3+1999=2002

3a^2-a-1=0
原式=6a^2+7a^2-5a+1999
=(6a^3-2a^2-2a)+(9a^2-3a-3)+1999+3
=2a(3a^2-a-1)+3(3a^2-a-1)+2002
=2002

6a³+7a²-5a+1999
=6a³-2a²-2a+9a²-3a-3+2002
=2a(3a²-a-1)+3(3a²-a-1)+2002
=0+0+2002
=2002
提示：就是通过裂项法构造因式3a²-a-1.