来自崔文华的问题
1.(3根号6-2根号1/6)-(根号24+2根号2/3)2.已知:a=2+根号5,b=2-根号5.求根号a²-ab+b²的值
1.(3根号6-2根号1/6)-(根号24+2根号2/3)
2.已知:a=2+根号5,b=2-根号5.求根号a²-ab+b²的值
1回答
2020-12-08 18:23
1.(3根号6-2根号1/6)-(根号24+2根号2/3)2.已知:a=2+根号5,b=2-根号5.求根号a²-ab+b²的值
1.(3根号6-2根号1/6)-(根号24+2根号2/3)
2.已知:a=2+根号5,b=2-根号5.求根号a²-ab+b²的值
1.(3根号6-2根号1/6)-(根号24+2根号2/3)
=(3√6-√6/3)-(2√6+2√6/3)
=0
2.已知:a=2+根号5,b=2-根号5.求根号a²-ab+b²的值
a=2+√5
b=2-√5
则a+b=4ab=-1
√(a²-ab+b²)=√=√(16+3)=√19