已知2x^2-3=0,求代数式x^2（x^2-1)^2+x^2(5-x^2)^2-9的值
问题描述：
已知2x^2-3=0,求代数式x^2（x^2-1)^2+x^2(5-x^2)^2-9的值
答
2x^2-3=0
x^2=3/2
x^2（x^2-1)^2+x^2(5-x^2)^2-9=3/2(3/2-1)^2+3/2(5-3/2)^2-9=3/8+147/8-9=78/8=39/4
答
39/4
答
解方程得x=3/4，代入到后面式子中求出值为3.
答
“^”这是啥？乘号还是。。
答
2x^2-3=0→x^2=3/2
代数式x^2（x^2-1)^2+x^2(5-x^2)^2-9
=3/2（3/2-1)^2+3/2(5-3/2)^2-9
=3/2（1/2)^2+3/2(7/2)^2-9
=3/2（1/4+49/4）-9
=3/2×25/2-9
=75/4-9
=39/4
答
2x²-3=0
2x²=3
x²=3/2
x²（x²-1)²+x²(5-x²)²-9
=x²[x^4-2x²+1+25-10x²+x^4]-9
=2x²(x^4-6x²+13)-9
=2x²[(x²-3)²+4]-9
=3(9/4+4)-9
=75/4-36/4
=39/4
答
9.75