Question

数学题，求解，详细过程 尽快

1.已知log 7 3=a,log 7 4=b,用a,b表示log 7 42
2.已知lgab=5,lga•lgb=6,求a,b的值
3.已知lga,lgb是方程x2+2x-4=0的两个根，求ab的值
4．已知x1,x2是方程x2-mx+3=0的两个根，且lg(x1+x2)/lgx1+lgx2=2,求m的值

Answer 1

1 ： log 7 4 = log 7 2^2 = 2log 7 2 = b
所以 log 7 2 = b/2
log 7 3 + log 7 2 + 1
= log 7 3*2 + log 7 7
= log 7 6*7
= log 7 42
即 a + b/2 + 1 = log 7 42
2： lgab = 5 = lga + lgb
即 lga = 5 - lgb
又因为 lga *lgb = 6
所以 (5 - lgb)lgb = 6
即 （lgb）^2 - 5lgb +6 =0
解得 lgb = 2或3
所以 当 lgb = 2时lga = 3 即 a = 1000 , b =100
当 lgb = 3时lga = 2 即 a = 100 , b =1000
3：由韦达定理可知
lga + lgb = -2
lga * lgb = -4
解得 lga = -1- 根号5 lgb = -1 + 根号5
或 lga = -1+ 根号5 lgb = -1 - 根号5
lga + lgb = lgab = -2
所以 ab = 1/100
4：由韦达定理可知
x1 + x2 = m
x1*x2 = 3
所以 lg(x1+x2)/(lgx1+lgx2)
= lgm/lg3
= 2
即 lgm = 2lg3 = lg3^2 = lg9
所以 m = 9

Answer 2

1)log7(42)=log7(3*14)=log7 (3)+log7(14)=log7(3)+log7(2)+log7(7)=a+b/2+1
2)lgab=lga+lgb=5,lga*lgb=6
把lga,lgb看作是方程x^2-5x+6=0的两个根
lga=2,lgb=3或lga=3,lgb=2
a=100,b=1000或b=100,a=1000
3)lga+lgb=-2
lgab=-2
ab=1/100
4)x1+x2=m,x1x2=3
原式:lgm/lg3=2
lgm=2lg3
lgm=lg9
m=9

Answer 3

1.log 7 4=b
∵log 7 4=log 7 2^2=2log 7 2=b
∴log 7 2=b/2
∴log 7 42=log 7 (6*7)=log 7 6+log 7 7=log 7 (2*3)+log 7 7
=log 7 2+log 7 3+log 7 7=b/2+a+1
2.lgab=5,lga•lgb=6
lgab=lga+lgb=5
lga=5-lgb
代入lga•lgb=6
有lgb(5-lgb)=6
5lgb-(lgb)^2-6=0
(lgb-2)(lgb-3)=
lgb=2或lgb=3
b=100或b=1000
当b=100时，a=1000
当b=1000时，a=100
3.已知lga,lgb是方程x2+2x-4=0的两个根，求ab的值
有lga+lgb=-2,
所以lga+lgb=lg(a*b)=-2,
所以ab=1/100
4.已知x1,x2是方程x2-mx+3=0的两个根，且lg(x1+x2)/lgx1+lgx2=2,求m的值
有x1+x2=m,x1x2=3
lg(x1+x2)=lgm
lgx1+lgx2=lgx1x2=lg3
lg(x1+x2)/lgx1+lgx2=lgm/lg3=2
所以m=9