你哪里不懂啊
a^2-10a+b-2√(b-4)+|√(c-1)-2|+22=0
则a^2-10a+25-25+b-2√(b-4)+|√(c-1)-2|+22=0
a^2-10a+25+b-2√(b-4)+|√(c-1)-2|-3=0
a^2-10a+25+(b-4)+4-2√(b-4)+|√(c-1)-2|-3=0
a^2-10a+25+(b-4)-2√(b-4)+1+|√(c-1)-2|=0
所以(a-5)^2+[√(b-4)-1]^2+|√(c-1)-2|=0
则a-5=0,√(b-4)-1=0,√(c-1)-2=0
解得a=b=c=5