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Did it with logs:3 log b = log b + log 32 log b = log 3log b = (log 3)/2b = √3This gives one solution, but doesn't really give any hints about the other two. Finding those requires some further thought.
b^3=3bb^2=3b=+/-sqrt(3)a 3rd solution is missingsob^3=3bb^3-3b=0b(b^2-3)=0b=0 sol.1b=+/-sqrt(3) sol.2, sol.3sqrt(3) = 1.7321.732^3=?3(1.732) =? 3•1.7321.732•1.732•1.732 =? 3•1.7323•1.732 = 3•1.733✔️b=? -1.732(-1.732)^3=3(-1.732)(-1.732)^2•(-1.732) =? 3(-1.732)3•(-1.732) =❤ 3•(-1.732)✔️
b³-3b=0b(b²-3)=0So roots are 0 and ±sqrt3
{b+b ➖ b+b➖ b+b ➖ }={b^2+b^2+b^2}=b^6 b^2^2^2 b^1^1^2 b^1^2 (b ➖ 2b+1).
b=0
since there's b^3 = 3bthere's 3 solutionsb=0 is 1 of 3 .. so you get 1/3 credit
Root 3
What about b=- Root 3 or b=0?
3root 3
Did it with logs:
3 log b = log b + log 3
2 log b = log 3
log b = (log 3)/2
b = √3
This gives one solution, but doesn't really give any hints about the other two. Finding those requires some further thought.
b^3=3b
b^2=3
b=+/-sqrt(3)
a 3rd solution is missing
so
b^3=3b
b^3-3b=0
b(b^2-3)=0
b=0 sol.1
b=+/-sqrt(3) sol.2, sol.3
sqrt(3) = 1.732
1.732^3=?3(1.732)
=? 3•1.732
1.732•1.732•1.732
=? 3•1.732
3•1.732 = 3•1.733✔️
b=? -1.732
(-1.732)^3=3(-1.732)
(-1.732)^2•(-1.732)
=? 3(-1.732)
3•(-1.732) =❤ 3•(-1.732)✔️
b³-3b=0
b(b²-3)=0
So roots are 0 and ±sqrt3
{b+b ➖ b+b➖ b+b ➖ }={b^2+b^2+b^2}=b^6 b^2^2^2 b^1^1^2 b^1^2 (b ➖ 2b+1).
b=0
since there's b^3 = 3b
there's 3 solutions
b=0 is 1 of 3 .. so you get 1/3 credit
Root 3
What about b=- Root 3 or b=0?
3root 3