作业帮求这四道题的答案（二次根式）
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/19 01:37:52
求这四道题的答案（二次根式）
要过程
解1题：
原式=√(3²×2)+1/(√2+1)+1/(-2)²
=3√2+(√2-1)/[(√2-1)(√2+1)]+1/4
=3√2+(√2-1)/(√2²-1)+1/4
=3√2+(√2-1)/(2-1)+1/4
=3√2+√2-1+1/4
=4√2-3/4
=(16√2-3)/4
解2题：
原式=4√5+√(3²×5)-√(2²×2)+4√2
=4√5+3√5-2√2+4√2
=7√5+2√2
解3题：
原式=6-5√(3/2)
=6-5√(6/4)
=6-5×(√6)/2
=(12-5√6)/2
解4题：
原式=2(√2+1)/[(√2+1)(√2-1)]+√(3²×2)-4√(2/4)
=2(√2+1)/(√2²-1)+3√2-4×(√2)/2
=2(√2+1)/(2-1)+3√2-2√2
=2(√2+1)+√2
=2√2+2+√2
=3√2+2
原式=√(3²×2)+1/(√2+1)+1/(-2)²
=3√2+(√2-1)/[(√2-1)(√2+1)]+1/4
=3√2+(√2-1)/(√2²-1)+1/4
=3√2+(√2-1)/(2-1)+1/4
=3√2+√2-1+1/4
=4√2-3/4
=(16√2-3)/4
解2题：
原式=4√5+√(3²×5)-√(2²×2)+4√2
=4√5+3√5-2√2+4√2
=7√5+2√2
解3题：
原式=6-5√(3/2)
=6-5√(6/4)
=6-5×(√6)/2
=(12-5√6)/2
解4题：
原式=2(√2+1)/[(√2+1)(√2-1)]+√(3²×2)-4√(2/4)
=2(√2+1)/(√2²-1)+3√2-4×(√2)/2
=2(√2+1)/(2-1)+3√2-2√2
=2(√2+1)+√2
=2√2+2+√2
=3√2+2