Найдите пятый член прогрессии если известно что b1=6, b3=2/3

b1=6  b3=2/3

b1*q^2=b3

6*q^2=2/3

q^2=2/18

q^2=1/9

q=+-1/3

q1=1/3                                                                                        q2=-1/3

b5(1)=b1*q^4=6*(1/3)^4=2/27                      b5(2)=b1*q^4=6*(-1/3)^4=2/27

 

ответ: 2/27

 

 

First, we'll try to plug in the value: #lim_{x to -oo}x+sqrt(x^2+2x) = -oo + sqrt(oo-oo)# we're already encountering a problem: it is simply not allowed to have #oo-oo#, it's like dividing by zero. we need to try a different approach. whenever i see this kind of limit, i try to use a trick: #lim_{x to -oo}x+sqrt(x^2+2x)# #= lim_{x to -oo}x+sqrt(x^2+2x)*(x-sqrt(x^2+2x))/(x-sqrt(x^2+2x))# these are the same becaus the factor we're multiplying with is essentially #1#. why are we doing this? because there exists a formula which says: #(a-b)(a+b) = a^2-b^2# in this case #a = x# and #b = sqrt(x^2+2x)# let's apply this formula: #lim_{x to -oo}(x^2-(sqrt(x^2+2x))^2)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(x^2-x^2-2x)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(-2x)/(x-sqrt(x^2+2x))# now we're going to use another trick. we'r going to use this one, because we want to get the #x^2# out of the square root: #lim_{x to -oo}(-2x)/(x-sqrt(x^2(1+2/x))# if you look carefully, you see it's the same thing. now, you might say that #sqrt(x^2) = x#, but you have to remember that #x# is a negative number. because we're taking the positive square root, #sqrt(x^2) = -x# in this case. #= lim_{x to -oo}(-2x)/(x+xsqrt(1+2/x))# #= lim_{x to -oo}(-2x)/(x(1+sqrt(1+2/# we can cancel the #x#: #= lim_{x to -oo}(-2)/(1+sqrt(1+2/x))# and now, we can finally plug in the value: #= -2/(1+sqrt(1+2/-oo))# a number divided by infinity, is always #0#: #= -2/(1+sqrt(1+0)) = -2/(1+1) = -2/2 = -1# this is the final answer. hope it helps.

4|x-2|  +  2x  =  3|x-2|  +  1 4|x-2|  +  2x -  3|x-2| =1 (4-3) * |x-2| + 2x = 1 1 * |x-2| + 2x = 1 |x-2| + 2x = 1   разделим уравнение на 2 возможных случаев : x - 2 + 2x = 1,  когда :     x - 2  ≥ 0 -(x-2) + 2x = 1, когда :   x - 2 < 0 x - 2 + 2x = 1,      x - 2  ≥ 0 -x  +  2 + 2x = 1,    x - 2 < 0 x + 2x = 1 + 2,    x  ≥ 2 -x + 2x = 1 - 2,    x < 2   3x = 3 | : 3,  x  ≥ 2 x = -1 | : 1,    x < 2 x = 1,    x  ≥ 2 x = -1,  x < 2 x  ∈  ∅ - т.к. x ≥ 2, x не будет 1 x = -1 - верно, т.к. x < 2, x = -1   ответ : x = -1