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/x)))# 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.
#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/x)))#
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.
Объяснение:
x1 - x2 + 2x3 = -2
x1 + 2x2 - x3 = 7
2x1 + x2 - 3x3 = 5
Перепишем систему уравнений в матричном виде и решим его методом Гаусса
1 -1 2 -2
1 2 -1 7
2 1 -3 5
от 2 строки отнимаем 1 строку, умноженную на 1; от 3 строки отнимаем 1 строку, умноженную на 2
1 -1 2 -2
0 3 -3 9
0 3 -7 9
2-ую строку делим на 3
1 -1 2 -2
0 1 -1 3
0 3 -7 9
к 1 строке добавляем 2 строку, умноженную на 1; от 3 строки отнимаем 2 строку, умноженную на 3
1 0 1 1
0 1 -1 3
0 0 -4 0
3-ую строку делим на -4
1 0 1 1
0 1 -1 3
0 0 1 0
от 1 строки отнимаем 3 строку, умноженную на 1; к 2 строке добавляем 3 строку, умноженную на 1
1 0 0 1
0 1 0 3
0 0 1 0
x1 = 1
x2 = 3
x3 = 0
Сделаем проверку. Подставим полученное решение в уравнения из системы и выполним вычисления:
1 - 3 + 2·0 = 1 - 3 + 0 = -2
1 + 2·3 - 0 = 1 + 6 + 0 = 7
2·1 + 3 - 3·0 = 2 + 3 + 0 = 5
Проверка выполнена успешно.
x1 = 1
x2 = 3
x3 = 0
ЕСЛИ НЕ ПОНЯТНО, ТО ВОТ ССЫЛКА:https://ru.onlinemschool.com/math/assistance/equation/gaus/