√ab+2√b/a+√1/ab-6√a/b(a>0,b>0)完整过程,急用- -- --- --√ab ﹢2√b/a ﹢√1/ab -6√a/b (a>0,b>0)
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![√ab+2√b/a+√1/ab-6√a/b(a>0,b>0)完整过程,急用- -- --- --√ab ﹢2√b/a ﹢√1/ab -6√a/b (a>0,b>0)](/uploads/image/z/10131974-62-4.jpg?t=%E2%88%9Aab%2B2%E2%88%9Ab%2Fa%2B%E2%88%9A1%2Fab-6%E2%88%9Aa%2Fb%28a%3E0%2Cb%3E0%29%E5%AE%8C%E6%95%B4%E8%BF%87%E7%A8%8B%2C%E6%80%A5%E7%94%A8%EF%BC%8D+%EF%BC%8D%EF%BC%8D+%EF%BC%8D%EF%BC%8D%EF%BC%8D+%EF%BC%8D%EF%BC%8D%E2%88%9Aab+%EF%B9%A22%E2%88%9Ab%EF%BC%8Fa+%EF%B9%A2%E2%88%9A1%EF%BC%8Fab+%EF%BC%8D6%E2%88%9Aa%EF%BC%8Fb+%28a%EF%BC%9E0%2Cb%EF%BC%9E0%29)
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√ab+2√b/a+√1/ab-6√a/b(a>0,b>0)完整过程,急用- -- --- --√ab ﹢2√b/a ﹢√1/ab -6√a/b (a>0,b>0)
√ab+2√b/a+√1/ab-6√a/b(a>0,b>0)完整过程,急用
- -- --- --
√ab ﹢2√b/a ﹢√1/ab -6√a/b (a>0,b>0)
√ab+2√b/a+√1/ab-6√a/b(a>0,b>0)完整过程,急用- -- --- --√ab ﹢2√b/a ﹢√1/ab -6√a/b (a>0,b>0)
原式=根号(ab)+(2\a)根号(ab)+(1\ab)根号(ab)--(6\b)根号(ab)
=[1+(2\a)+1\ab)--6b]根号(ab)
=[(ab+2b+1--6ab^2)\(ab)]根号(ab).
问题的补充与原来的不是一样的吗?
化简[√ab-ab/(a+√ab)]÷(√ab-a)/(a-b)
计算[(√ab ) -ab/(a+√ab)]÷(√ab)-b/a-b
(√ab-ab/a+√ab)×√ab-b/a-b
计算:√(ab)×2√(b/a)×(-√a/b)(-√1/ab)
判断√ab和2ab/a+b的大小
>=a+b>=2√ab,
推导a+b>=2√ab
证明√(a^2+1/(b^2)+a^2/(ab+1)^2)=|a+1/b-a/(ab+1)|
计算题[a+2√(ab) +b]/(a-b)-[a+b-2√(ab)]/(√a-√b)
化简(√a+√b+√ab)^2-(√a+√b-√ab)^2
初三高一数学题一道.[√a+(b-√ab)/(√a+√b)]÷[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]
[√a+(b-√ab)/(√a+√b)]/[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]=
a√(a/b)÷√ab×√(1/ab)(b>0)
a√a/b÷√ab×√1/ab(b大于0)
(a- √ab -2b)分之(a - 2√ab) +(a√a + b√b)分之(a√b -b√a +b√b) 化简
先化简,再求值:(A^2-2/A^2-AB^2+2-A^2B-AB^2)/(A^2+AB/2AB+B^2+ab/2ab),其中A=5-√11,B=3+√11
关于高中基本不等式若正数A,B满足AB=A+B+3,则AB的取值范围是:AB=A+B+3≥2√AB+3AB-2√AB-3≥0(√AB-3)(√AB+1)≥0(√AB)≥3AB)≥9(当且仅当A=B时取等号)以上没有问题疑惑:利用基本不等式 ab=0ab>=9或ab
高二不等式比较大小已知f(x)=(1+√(1+x))/x,a、b是两个不相等的实数,则下列不等式正确的是( )A.f((a+b)/2)>f(√ab)>f(2ab/(a+b)B.f((a+b)/2)>f(2ab/(a+b)>f(√ab)C.f(2ab/(a+b)>f(√ab)>f((a+b)/2)D.f(√ab)>f(2ab/(a+b)>f((a+