宇宙级数学难题已知x^(0.5)+1/x^(0.5)=3,求(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)的值,可得3分追加分
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/14 04:09:08
![宇宙级数学难题已知x^(0.5)+1/x^(0.5)=3,求(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)的值,可得3分追加分](/uploads/image/z/12251519-71-9.jpg?t=%E5%AE%87%E5%AE%99%E7%BA%A7%E6%95%B0%E5%AD%A6%E9%9A%BE%E9%A2%98%E5%B7%B2%E7%9F%A5x%5E%280.5%29%2B1%2Fx%5E%280.5%29%3D3%2C%E6%B1%82%28x%5E%283%2F2%29%2Bx%5E%28-3%2F2%29%2B2%29%2F%28x%5E2%2Bx%5E%28-2%29%2B3%29%E7%9A%84%E5%80%BC%2C%E5%8F%AF%E5%BE%973%E5%88%86%E8%BF%BD%E5%8A%A0%E5%88%86)
宇宙级数学难题已知x^(0.5)+1/x^(0.5)=3,求(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)的值,可得3分追加分
宇宙级数学难题
已知x^(0.5)+1/x^(0.5)=3,求(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)的值,可得3分追加分
宇宙级数学难题已知x^(0.5)+1/x^(0.5)=3,求(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)的值,可得3分追加分
我来解答一下吧:把已知条件变为:x^(0.5)+x^(-0.5)=3
首先解分母:x^2+x^(-2)+3=[x+x^(-1)]^2-2+3=[x+x^(-1)]^2+1
而x+x^(-1)=[x^(0.5)+x^(-0.5)]^(2)-2=7
所以由上式可得分母的值:7^(2)+1=50
下面再来做分子:x^(3/2)+x^(-3/2)=[x^(0.5)]^3+[x^(-0.5)]^3=[x^(0.5)+x^(-0.5)][x+x^(-1)-1]=3*6=18
所以分子为20,所以最后原式的值:0.4即5分之2
解:
(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)
=(x^(3/2)+1/x^(3/2)+2)/(x^2+1/x^2+3)
利用立方和公式
=[(x^1/2+1/x^1/2)*(x+1/x -1)+2]/[(x+1/x)^2+1]
因为:x^(0.5)+1/x^(0.5)=3
所以:x+1/x=3^2-2=7
=3*(x+1/x -1)+2]/[(x+1/x)^2+1]
=20/50
=2/5
=12/25
x^(0.5)+1/x^(0.5)=3
[x^(0.5)+1/x^(0.5)]^2=3^2
x+2+x^(-1)=9
x+x^(-1)=7
[x^(3/2)+x^(-3/2)+2]/[x^2+x^(-2)+3]
={[x^1/2+x^(-1/2)]*[x-1+x^(-1)]+2}/{[x+x^(-1)]^2+1}
=[3*(7-1)+2]/50
=20/50
2/5
x^(0.5)+1/x^(0.5)=3则x+1/x=3×3-2=7,进而x^2+x^(-2)+3=7×7-2+3=50
(x+1/x)×[x^(0.5)+1/x^(0.5)]=x^(3/2)+x^(-3/2)+x^(0.5)+1/x^(0.5)=x^(3/2)+x^(-3/2)+3=7×3=21,所以x^(3/2)+x^(-3/2)+2=21-3+2=20
原式=20/50=0.4。