Question

(1)已知实数x,y,z满足x=6-y,z^2=xy-9,求证:x=y (2)已知x^2+x-1=0,求代数式21x^4-28x^3-70x^2+49x+1

快

Answer 1

第一题啊;x=6-y带入第二个得到z^2=-(3-y)^2,那么y=3,那么x=3
第二题：21x^4-28x^3-70x^2+49x+1=21x^4+21x^3-49x^3-.....
=21x^2-70x(x^2+x)+21x^3+49x+1
=21x-70x+49x+1
=1

Answer 2

（1）实数x,y,z满足x=6-y,z^2=xy-9=（6-y)y-9=-(y-3)^2>=0,
∴y=3,x=3,
∴x=y.
(2)21x^4-28x^3-70x^2+49x+1 |x^2+x-1
....21x^4+21x^3-21x^2............----------
--------------------------............21x^2-49x
............-49x^3-49x^2+49x
............-49x^3-49x^2+49x
..........--------------------------
.........................................1,
∴21x^4-28x^3-70x^2+49x+1
=（x^2+x-1）（21x^2-49x)+1
=1.

Answer 3

(1)证明：x=6-y
z^2=xy-9
=(6-y)y-9
=-(y^2-6y+9)
=-(y-3)^2
x,y,z是实数，所以z=0，y=3
x=6-y=3
所以x=y
（2）x^2+x-1=0 x^2+x=1
21x^4-28x^3-70x^2+49x+1
=7×(3x^4-4x^3-10x^2+7x)+1
=7×{3x^2(x^2+x)-7x(x^2+x)-3(x^2+x)+10x}+1
=7×(3x^2-7x-3+10x)+1
=7×(3x^2+3x-3)+1
=7×3+1
=22