Analysis: x-1≥0, x≥1
Solution ①: Let ³√(2-x)=a, √(x-1)=b
∴a=1-b…①
a³+b²=1…②
0a³+b²=1…②
0a³+b²=1…① ②
a³+(1-a)²=1
∴a³+a²-2a=0
∴a( a²+a-2)=0
∴a(a+2)(a-1)=0
∴a=0 or a=-2 or a=1
When a=0, ³√(2-x)=0, x=2
When a=-2, ³√(2-x)=-2, x=10
When a=1, ³√(2-x)=1, x=1
∴Experiential root, the solution of the original equation is: x1=2, x2=10, x3=1
When a=1, ³√(2-x)=1, x=1
∴Experiential root, the solution of the original equation is: x1=2, x2=10, x3=1
∴Experiential root, the solution to the original equation is: x1=2, x2=10, x3=1
∴Experiential root, the solution to the original equation is: x1=2, x2=10, x3=1
∴Experiential root, the solution to the original equation is: x1=2, x2=10, x3=1
html l3 solution ②: Let ³√(2-x)=a=1-√(x-1)
∴³√(2-x)=a…①
1-√(x-1)=a…②
from ①:2-x=a³,∴x=2-a³…③
from ②:√(x-1)=1-a
from ②:√(x-1)=1-a
∴x-1=a²-2a+1,∴x=a²-2a+2…④
from ③,④:2-a³=a²-2a+2
htt ml3∴a³+a²-2a=0
∴a(a+2)(a-1)=0
∴a=0 or a=-2 or a=1
When a=0, ³√(2-x)=0, ∴x=2
When a=-2, ³√(2-x)=-2, ∴x=10
When a=0, ³√(2-x)=1, ∴x=1
When a=0, ³√(2-x)=1, ∴x=1
∴empirical root, the solution of the original equation is: x1=2, x2=10, x3=1
