求20道二次根式的计算题附答案盒过程,不要知道里有过的!
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/07 19:33:58
求20道二次根式的计算题附答案盒过程,不要知道里有过的!
不要填空选择啥的= =20悬赏哦·····好的话可以追赏
不要填空选择啥的= =20悬赏哦·····好的话可以追赏
1.X(2-√2)=(√2-2)
X(2-√2)=(√2-2)
X(2-√2)=-1(2-√2)
x=1
2.(√5+√2)²-(√5-√2)²
=(√5+√2+√5-√2)(√5+√2-√5+√2)
=2√5*2√2
=4√10
3.√8+3√(1/3)-1/(√2)+(√3)/2
=2√2+3*(√3)/3-(√2)/2+(√3)/2
=2√2+√3-(√2)/2+(√3)/2
=(3/2)√3+(3/2)√2
4.(√3+√2+√5)(√3-√2-√5)
=(√3+√2+√5)[√3-(√2+√5)]
=(√3)²-(√2+√5)²
=3-(2+5+2√10)
=3-7-2√10
=-4-2√10
5.√8-2√32+√50
=5*3√2-2*4√2+5√2
=√2(15-8+5)
=12√2
6.√6-√3/2-√2/3
=√6-√6/2-√6/3
=√6/6
7.(√45+√27)-(√4/3+√125)
=(3√5+3√3)-(2√3/3+5√5)
=-2√5+7√5/3
8.(√4a-√50b)-2(√b/2+√9a)
=(2√a-5√2b)-2(√2b/2+3√a)
=-4√a-6√2b
9.(√32-3√3)(4√2+√27)
=(4√2-3√3)(4√2+3√3)
=(4√2)^2-(3√3)^2
=32-27
=5
10.1+√2-√3)(1-√2+√3)
=[1+(√2-√3)][1-(√2-√3)]
=1-(√2-√3)^2
=1-(2+3+2√6)
=-4-2√6
美女,10道够不够啊...
X(2-√2)=(√2-2)
X(2-√2)=-1(2-√2)
x=1
2.(√5+√2)²-(√5-√2)²
=(√5+√2+√5-√2)(√5+√2-√5+√2)
=2√5*2√2
=4√10
3.√8+3√(1/3)-1/(√2)+(√3)/2
=2√2+3*(√3)/3-(√2)/2+(√3)/2
=2√2+√3-(√2)/2+(√3)/2
=(3/2)√3+(3/2)√2
4.(√3+√2+√5)(√3-√2-√5)
=(√3+√2+√5)[√3-(√2+√5)]
=(√3)²-(√2+√5)²
=3-(2+5+2√10)
=3-7-2√10
=-4-2√10
5.√8-2√32+√50
=5*3√2-2*4√2+5√2
=√2(15-8+5)
=12√2
6.√6-√3/2-√2/3
=√6-√6/2-√6/3
=√6/6
7.(√45+√27)-(√4/3+√125)
=(3√5+3√3)-(2√3/3+5√5)
=-2√5+7√5/3
8.(√4a-√50b)-2(√b/2+√9a)
=(2√a-5√2b)-2(√2b/2+3√a)
=-4√a-6√2b
9.(√32-3√3)(4√2+√27)
=(4√2-3√3)(4√2+3√3)
=(4√2)^2-(3√3)^2
=32-27
=5
10.1+√2-√3)(1-√2+√3)
=[1+(√2-√3)][1-(√2-√3)]
=1-(√2-√3)^2
=1-(2+3+2√6)
=-4-2√6
美女,10道够不够啊...