已知lga,lgb是方程
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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
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
lga+lgb=lg(2a+b),lg(ab)=lg(2a+b)ab=2a+b≧2√(2ab)即:ab≧2√(2ab)a²b²≧8ab得:ab≧8当且仅当2a=b时等号成立.祝你开
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
都是正数所以a+b>=2√abb+c>=2√bcc+a>=2√ca相乘(a+b)(b+c)(c+a)>=8√(a^2b^2c^2)即(a+b)(b+c)(c+a)>=8abc要取等号则上面三个式子的等
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-
把条件转化为ab=1,∴b1+a2+a1+b2=b2b+a2b+a2a+ab2=b2b+a+a2a+b =a2+b2a+b=2(a2+b2)2(a+b)≥a2+b2+2ab2(a+b)=(a
x=√(lga+1)+√(lgb+1)则显然x>=0ab=1000lg(a+b)=lg1000lga+lgb=3x²=lga+lgb+2+2√(lgalgb+lga+lgb+1)=5+2√(