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.
x^2 ≤ 7x + 60 Запишем равенство и найдём корни квадратного уравнения. x^2 - 7x - 60 = 0 D = 7 * 7 + 4 * 60 = 289 = 17^2 x1.2 = (7 ± 17)/2 x1 = 24/2 = 12 x2 = - 10/2 = - 5 Разложим левую часть неравенства на множители. (x − 12)(x + 5) ≤ 0 Произведение двух сомножителей отрицательно в том случае, когда один из них отрицательный, а второй положительный. Поэтому можем записать две системы неравенств. x - 12 ≤ 0 x + 5 ≥ 0 и x - 12 > 0 x + 5 ≤ 0 x ≤ 12 x ≥ - 5 [- 5; 12] - интервал значений переменной, удовлетворяющих неравенству. x ≥ 12 x ≤ - 5 Найдём длину полученного интервала 12 + 5 = 17 единиц. ответ: длина интервала, на котором выполняется неравенство, 17 единиц.
#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.
В)17
Объяснение:
x^2 ≤ 7x + 60 Запишем равенство и найдём корни квадратного уравнения. x^2 - 7x - 60 = 0 D = 7 * 7 + 4 * 60 = 289 = 17^2 x1.2 = (7 ± 17)/2 x1 = 24/2 = 12 x2 = - 10/2 = - 5 Разложим левую часть неравенства на множители. (x − 12)(x + 5) ≤ 0 Произведение двух сомножителей отрицательно в том случае, когда один из них отрицательный, а второй положительный. Поэтому можем записать две системы неравенств. x - 12 ≤ 0 x + 5 ≥ 0 и x - 12 > 0 x + 5 ≤ 0 x ≤ 12 x ≥ - 5 [- 5; 12] - интервал значений переменной, удовлетворяющих неравенству. x ≥ 12 x ≤ - 5 Найдём длину полученного интервала 12 + 5 = 17 единиц. ответ: длина интервала, на котором выполняется неравенство, 17 единиц.