已知lg2=a,lg3=b,则log3(6)a+b/aa+b/ba/a+bb/a+b(a+b)/a(a+b)/ba/(a+b)b/(a+b)

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已知lg2=a,lg3=b,则log3(6)a+b/aa+b/ba/a+bb/a+b(a+b)/a(a+b)/ba/(a+b)b/(a+b)

已知lg2=a,lg3=b,则log3(6)a+b/aa+b/ba/a+bb/a+b(a+b)/a(a+b)/ba/(a+b)b/(a+b)
已知lg2=a,lg3=b,则log3(6)
a+b/a
a+b/b
a/a+b
b/a+b
(a+b)/a
(a+b)/b
a/(a+b)
b/(a+b)

已知lg2=a,lg3=b,则log3(6)a+b/aa+b/ba/a+bb/a+b(a+b)/a(a+b)/ba/(a+b)b/(a+b)
log3(6)
= log3(2×3)
= log3(2) + log3(3)
= log3(2) + 1
= lg2/lg3 + 1
= a/b + 1
= (a + b)/b
所以选 第二个

因为lg2=a,lg3=b,
由换底公式得
log(3)(6)=lg6/lg3
=(lg2+lg3)/lg3
=(a+b)/b.
选B.

笨,这都不会?晕