3二次根式的题目~~~有分加~~~过程啊~~~1.a^2+6b-4ab-ab+2=0,计算 √a+√b/√(3a)的值2.化简 √下x^2+1/(x)^2-2 (0
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 20:13:50
![3二次根式的题目~~~有分加~~~过程啊~~~1.a^2+6b-4ab-ab+2=0,计算 √a+√b/√(3a)的值2.化简 √下x^2+1/(x)^2-2 (0](/uploads/image/z/7060488-24-8.jpg?t=3%E4%BA%8C%E6%AC%A1%E6%A0%B9%E5%BC%8F%E7%9A%84%E9%A2%98%E7%9B%AE%7E%7E%7E%E6%9C%89%E5%88%86%E5%8A%A0%7E%7E%7E%E8%BF%87%E7%A8%8B%E5%95%8A%7E%7E%7E1.a%5E2%2B6b-4ab-ab%2B2%3D0%2C%E8%AE%A1%E7%AE%97++%E2%88%9Aa%2B%E2%88%9Ab%2F%E2%88%9A%EF%BC%883a%EF%BC%89%E7%9A%84%E5%80%BC2.%E5%8C%96%E7%AE%80+%E2%88%9A%E4%B8%8Bx%5E2%2B1%2F%28x%29%5E2-2+%280)
xQN@.ԡ@|!4LLiZb"NY>_ɝs9wZ.ȝDgtŰ^G6or{8~mim@
2TQJq'&»`fJt)RΕ$'\tL5T$uvKZKBiWZU_;
lzt; _W@#\P²;_[=W pLbD_[5Se#iMJ 3TQ LInK ӵ=S'\i>q{
'nöO. _8 ^:]4r,\ts%
psR.]wd
3二次根式的题目~~~有分加~~~过程啊~~~1.a^2+6b-4ab-ab+2=0,计算 √a+√b/√(3a)的值2.化简 √下x^2+1/(x)^2-2 (0
3二次根式的题目~~~有分加~~~过程啊~~~
1.a^2+6b-4ab-ab+2=0,计算 √a+√b/√(3a)的值
2.化简 √下x^2+1/(x)^2-2 (0
3二次根式的题目~~~有分加~~~过程啊~~~1.a^2+6b-4ab-ab+2=0,计算 √a+√b/√(3a)的值2.化简 √下x^2+1/(x)^2-2 (0
1.题目条件应当有误
2.根号〔(1/(x)^2)(x^4+1-2x^2)〕=(1/x)根号[(x^2-1)^2]
=(1/x)(1-x^2)=(1/x)-x
3.a^2+2ab+b^2=(a+b)^2=(2√3)^2=8
你应该自己动动脑啊,像是第三题和第二题根本就不用问的啊
根号〔(1/(x)^2)(x^4+1-2x^2)〕=(1/x)根号[(x^2-1)^2]
=(1/x)(1-x^2)=(1/x)-x
3. a^2+2ab+b^2=(a+b)^2=(2√3)^2=8