(lg2)^3+3lg2lg5+(lg5)^3 求值
发布网友
发布时间：1天前
共0个回答
(lg2)^3+(lg5)^+3lg2lg5 =(lg2+lg5)^3-3(lg2)^2lg5-3lg2(lg5)^2+3lg2lg5 =(lg(2*5))^3-3lg2lg5(lg2+lg5-1)=1^3-3lg2lg5(lg10-1)=1
(lg2)^3+3lg2·lg5+(lg5)^3 怎么算??求详细过程和为什么~~＝(lg2)^3+(lg5)^3+3lg2·lg5 =(lg2+lg5)[(lg2)^2+(lg5)^2-lg2lg5]+3lg2·lg5 =[(lg2+lg5)^2-3lg2lg5]+3lg2·lg5 =1
(lg2)³+(lg5)³+3lg2×lg5(等于多少?)令a=lg2;b=lg5; 易得：a+b=1; 原式子可化解为： a^3+b^3+3*a*b=(a+b)(a^2+b^2-ab)+3ab =a^2+b^2-ab+3ab=a^2+b^2+2ab=(a+b)^2=1
请问这道数学题怎么做?(lg2)³+(lg5)³+3lg2×lg5=?答案等于1哦，告诉你一个公式 a^3+b^3=(a+b)(a^2-ab+b^2),代入做就可以
(lg2)^3+(lg5)^3+3lg2*lg5 这些怎么算.令a=lg2,b=lg5 则a+b=lg2+lg5=1 原式=a³+b³+3ab =(a+b)(a²-ab+b²)+3ab =a²-ab+b²+3ab =a²+2ab+b²=(a+b)²=1
(lg2)^3+3lg2*1g5+(lg5)^3等于多少因为（a+b）^3 = a^3+3a^2b+3ab^2+b^3 所以原式 = （lg2+lg5）^3 - 3(lg2)^2lg5+3lg2*(1g5)^2+3lg2*1g5 = 1 -3lg2 - 3lg5 +3lg2*1g5 = 1 -3(lg2+lg5)+3lg2*lg5 = 1
lg2的3次方+3lg2×lg5+lg5的3次方(lg2)³+（lg5)³+3lg2×lg5 =(lg2+lg5)[(lg2)²-lg2×lg5+(lg5)²]+3lg2×lg5 (这步是用立方和公式)=(lg2)²-lg×lg5+(lg5)²+3lg2×lg5 =(lg2)²+2lg2*lg5+(lg5)²=（lg2+lg5）²=1 ...
求值:(lg^2)^3+(lg^5)^3+3lg^2*lg^5令a=lg2,b=lg5,则a+b=lg10=1 (lg^2)^3+(lg^5)^3+3lg^2*lg^5 =a^3+b^3+3ab =(a+b)(a^2-ab+b^2)+3ab =a^2-ab+b^2+3ab =a^2+2ab+b^2 =(a+b)^2 =1 2^(1+1/2log2^5)=2^(1+log2^5^(1/2))=2^1*2^log2^5^(1/2))=2*5^(1/2)=2√5...
(lg2)³+(lg5)³+3×lg2×lg5这道题怎么算,谢谢看图片。首先用立方和公式将两个三次方化成两个整式相乘。再进行合并同类项，得到完全平方公式，再根据lg2+lg5=1化简即可。祝学习进步。
(lg2)^3+(lg5)^3+3lg2*lg5怎么算=(lg2+lg5)((lg2)^2-lg2*lg5+(lg5)^2)+3lg2*lg5 =lg(2*5)((lg2)^2-lg2*lg5+(lg5)^2)+3lg2*lg5 =(lg2)^2-lg2*lg5+(lg5)^2+3lg2*lg5 =(lg2)^2+2lg2*lg5+(lg5)^2 =(lg2+lg5)^2 =(lg(2*5))^2 =1 ...
