已知lg a,lg b是方程2x^2-4x 1=0的两个根

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

在方程ax²+bx+c=0中,其两根之差x1-x2=√(b²-4ac)/a则本题：(lgb/a)²=(lgb-lga)²=(x1-x2)²=[(-4)

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

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

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=

由跟与系数的关系可得：lga+lgb=-2（两根之和等于一次项的负系数）整理上式得：lga+lgb=lga*b=-2所以a*b=10^-2=1/100=0.01

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

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

2x²-4x+1=0,则：lga+lgb=2；（lga）*（lgb）=1/2∴lg（ab）=lga+lgb=2；∴（lga/b）²=（lga-lgb）²=（lga+lgb

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

lga+lgb=2lg(a-2b),lgab=lg(a-2b)^2ab=(a-2b)^2=a^2-4ab+4b^2a^2-5ab+4b^2=0(a-b)(a-4b)=0a=b,a=4b定义域a>0,b

（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-