(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
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(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
0
要过程给我留言
(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
=(1+lg3-1g240)/(1-lg9+lg36/5)+1
=(1-lg80)/(1-lg5/4)+1
=-1g8/lg8+1
=-1+1
=0
(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)+1
高一数学计算(1+1/2lg9-lg240)/(1-2/3lg27+lg36/5)
lg9*lg11-1=?
求值 100^(1/2lg9-lg2)=
证明(lg9)^2大于lg8.1
(1+1/2log9-lg240)/(1-2/3lg27+lg(36/5))+1=?
高中数学对数运算1.(根号2减1)的x次方等于2,求,x2.若log以2为底x加log以2为底y等于2,则x+y的最小值为?3.分子1+1/2lg9-lg240除以分母1-2/3lg27+lg36/5等于多少?4.lg4+lg5lg20+(lg5)^2等于多少?5.log以3为底的4
我急用.(不会的别答)全都要有过程...1.(lg根号27+lg8-3lg根号10)/lg1.2=_______2.4lg2+3lg5-lg1/5=__________3.(1+0.5lg9-lg240)/(1-2/3 lg27+lg36/5)=__________4.已知函数f(x)=loga【a在下面】(x+1),g(x)=loga{a在下面}(1-x)(a>0
lg4+lg9+2√(lg6)2-lg36+1=?
根号下lg²3 -lg9 +1
lg9 lg11 与1的大小关系A:lg9 lg11 >1B:lg9 lg11 =1C:lg9 lg11 <1请证明
(10*100^1/2lg9-lg2)+100^1/2-lg4根号410*(100^1/2lg9-lg2)+100^(1/2-lg4根号4)=?
计算:(1)5^log1/5 2 (2)100^1/2lg9-lg2
lg4+lg9+2*根号下(lg6)^2-2lg6+1=?
计算lg4+lg9+2倍的根号下(lg6)的2次-lg36+1
{√[(lg3)^2-lg9+1] *(lg√27+lg8-lg√1000)}/(lg0.3*lg1.2)
1-√ ( lg平方3 - lg9 +1)=
lg4+lg9+2根号下(lg6)²-lg36+1=?