B4=b1•q^3
1•q^3=1/8
q^3=1/8
q=1/2
b9=b1•q^8=1•1/256=1/256
b1•(q^6–1) -63/64
S6 = --------------- = ----------- =(63•2)/64
q–1 -1/2
= 63/32
2x^2-5x-4=0 |:2
x^2-2,5x-2=0
По теореме Виета,
x1+x2=2,5,
x1*x2=-2;
1) 1/x1^2+1/x2^2=(x1^2+x2^2)/(x1^2*x2^2)=((x1+x2)^2-2*x1*x2)/(x1*x2)^2=(6,25-2*(-2))/4=(6,25+4)/4=10,25/4=2,5625
2) x1*x2^4+x2*x1^4=x1*x2(x1^3+x2^3)=x1*x2(x1+x2)(x1^2-x1*x2+x2^2)=x1*x2(x1+x2)((x1+x2)^2-3*x1*x2)=2,5(-2)(6,25-3*2)=-5(6,25-6)=-5*0,25=-1,25
Ответ: 1) 2,5625; 2) -1,25
Я же решил его)
lg²x - lgx² = lg²3 - 1
lg²x - 2lgx + 1 = lg²3
(lgx - 1)² = lg²3
lgx - 1 = ±lg3
x = 10^(1 ± lg3)