若x^?+x^-?=9,则x^2+1/x^2=?若x^?-x^-?=3,则x^2+1/x^2=?我打错了,应是若x^1/2+x^-1/2=9,则x^2+1/x^2=?若x^1/2-x^-1/2=3,则x^2+1/x^2=?
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![若x^?+x^-?=9,则x^2+1/x^2=?若x^?-x^-?=3,则x^2+1/x^2=?我打错了,应是若x^1/2+x^-1/2=9,则x^2+1/x^2=?若x^1/2-x^-1/2=3,则x^2+1/x^2=?](/uploads/image/z/5255050-58-0.jpg?t=%E8%8B%A5x%5E%3F%2Bx%5E-%3F%3D9%2C%E5%88%99x%5E2%2B1%2Fx%5E2%3D%3F%E8%8B%A5x%5E%3F-x%5E-%3F%3D3%2C%E5%88%99x%5E2%2B1%2Fx%5E2%3D%3F%E6%88%91%E6%89%93%E9%94%99%E4%BA%86%EF%BC%8C%E5%BA%94%E6%98%AF%E8%8B%A5x%5E1%2F2%2Bx%5E-1%2F2%3D9%2C%E5%88%99x%5E2%2B1%2Fx%5E2%3D%3F%E8%8B%A5x%5E1%2F2-x%5E-1%2F2%3D3%2C%E5%88%99x%5E2%2B1%2Fx%5E2%3D%3F)
若x^?+x^-?=9,则x^2+1/x^2=?若x^?-x^-?=3,则x^2+1/x^2=?我打错了,应是若x^1/2+x^-1/2=9,则x^2+1/x^2=?若x^1/2-x^-1/2=3,则x^2+1/x^2=?
若x^?+x^-?=9,则x^2+1/x^2=?若x^?-x^-?=3,则x^2+1/x^2=?
我打错了,应是若x^1/2+x^-1/2=9,则x^2+1/x^2=?
若x^1/2-x^-1/2=3,则x^2+1/x^2=?
若x^?+x^-?=9,则x^2+1/x^2=?若x^?-x^-?=3,则x^2+1/x^2=?我打错了,应是若x^1/2+x^-1/2=9,则x^2+1/x^2=?若x^1/2-x^-1/2=3,则x^2+1/x^2=?
这就看你对问题的把握了,应定要仔细观察题目已知与未知的联系.本题中已知为一次,未知为二次,正好通过平方求得,
x^(1/2)+x^(-1/2)=9
x^2+1/x^2
=(x+1/x)^2-2
=[(x^(1/2)+x^(-1/2)^2-2]^2-2
=(81-2)^2-2
=6
会了第一个,第二个应该也会了吧,这是两次平方而已
x^(1/2)-x^(1/2)=3
x^2+1/x^2
=(x-1/x)^2+2
=[(x^(1/2)-x^(-1/2))^2+2]^2+2
=(3^2+2)^2+2
=121+2
=123
x^(1/2)+x^(-1/2)=9
x^2+1/x^2
=(x+1/x)^2-2
=[(x^(1/2)+x^(-1/2)^2-2]^2-2
=(81-2)^2-2
=6239
x^(1/2)-x^(1/2)=3
x^2+1/x^2
=(x-1/x)^2+2
=[(x^(1/2)-x^(-1/2))^2+2]^2+2
=(3^2+2)^2+2
=121+2
=123
望请采纳
x^(1/2)+x^(-1/2)=9
x^2+1/x^2
=(x+1/x)^2-2
=[(x^(1/2)+x^(-1/2)^2-2]^2-2
=(81-2)^2-2
=6239
x^(1/2)-x^(1/2)=3
x^2+1/x^2
=(x-1/x)^2+2
=[(x^(1/2)-x^(-1/2))^2+2]^2+2
=(3^2+2)^2+2
=121+2
=123
请采纳
x^(1/2)+x^(-1/2)=9
x^2+1/x^2
=(x+1/x)^2-2
=[(x^(1/2)+x^(-1/2)^2-2]^2-2
=(81-2)^2-2
=6239
x^(1/2)-x^(1/2)=3
x^2+1/x^2
=(x-1/x)^2+2
=[(x^(1/2)-x^(-1/2))^2+2]^2+2
=(3^2+2)^2+2
=121+2
=123