lga,lgb是方程3x2-2x-2 3=0的根

根据韦达定理,得lga+lgb=2lga·lgb=1/2∴(lga/b)²=(lga-lgb)²=(lga)²+(lgb)²-2lga·lgb=(lga+lgb

由题意得lga+lgb＝1，①lga•lgb＝m，②lg2a+4(1+lga)＝0，③由③得（lga+2）2=0，∴lga=-2，即a=1100④④代入①得lgb=1-lga=3，∴b=1000．⑤④

由题意知a≥0,b≥0,故：若a＞1,b＞1则（lga+lgb）＞1/2（lga+lgb）＞lg（a+b/2）若0≤a＜1,0≤b＜1则（lga+lgb）＜1/2（lga+lgb）＜lg（a+b/2）

(lgb分之a)²=[lg(a/b)]²=(lga-lgb)²=(lga)²-2(lga)(lgb)+(lgb)²=(lga)²+2(lga

因为lga+lgb=2,lga*lgb=1/2,所以(lg(b/a))^2=(lgb-lga)^2=(lgb+lga)^2-4lgb*lga=4-2=2

由韦达定理得：lga+lgb=4/2=2lgalgb=1/2因此lgalgblg(ab)=lgalgb(lga+lgb)=1/2*2=1

答：lga和lgb是方程2x^2-4x+1=0的两个根根据韦达定理有：lga+lgb=-(-4)/2=2(lga)*(lgb)=1/2所以：lg(ab)=2lg(a/b)^2=(lga-lgb)^2=

lga,lgb是方程2x方-4x+1=0的两实数根则lg(ab)=韦达定理得：lga+lgb=4/2=2lga*lgb=1/2lg(ab)=lga+lgb=2

lg(a/b)=lga-lgblg(a/b))^2=(lga-lgb)^2=（lga+lgb)^2-4lga*lgb=2^2-4*1/2=2选A

(lgb分之a)²=[lg(a/b)]²=(lga-lgb)²=(lga)²-2(lga)(lgb)+(lgb)²=(lga)²+2(lga

因为lga、lgb是方程x2-4x+1=0的两个根,所以lga+lgb=4,lga*lgb=1从而[lg(a/b)]^2=(lga-lgb)^2=(lga+lgb)^2-4lga*lgb=4^2-4*

m=9lg(x1+x2)=2lg(x1x2)x1+x2=(x1x2)^2由韦达定理x1+x2=mx1x2=3所以m=9

lga+lgb=2,lgalgb=1/2(lga/b)^2=(lga-lgb)^2=(lga+lgb)^2-4lgalgb=2

由已知,lga+lgb=4/2=2,lga*lgb=-1/2,因此[lg(a/b)]^2=(lga-lgb)^2=(lga+lgb)^2-4lga*lgb=2^2-4*(-1/2)=6.再问：可是答案

因为2x^2-4x+1=0所以lga+lgb=2,lgaxlgb=1/2则lg(ab)x(logab+logba)=2x(lga+lgb)^2-2lgaxlgb/lgab=2x2/3=3

（lgb/a）^2=(lgb-lga)^2=[lga+lgb]^2-4lga*lgb=4*4-2*1=14lga=

第一题：由题意有：lga+lgb=2lga*lgb=1/2.*又有(lga/b)^2=(lga-lgb)^2=(lga)^2+(lgb)^2-2lgalgb由*可得（lga+lgb)^2=(lga)^

=lga-lgb=√(lga-lgb)^2=√[(lga+lgb)^2-4lga*lgb]=√(4-4*1／2)=√2

(lga/b)²=(lga-lgb)²=(lga+lgb)²-4lga·lgb=2²-4×(1/2)=2你过程没错...再问：那答案错了我的是8真确是2为什么再

由韦达定理:lga+lgb=2,lga×lgb=1/2lg²(a/b)=(lga-lgb)²=(lga+lgb)²-4lga×lgb=2²-4×(1/2)=4-