плиз помогите первое и. третье задание решить ((
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Ответил m11m
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1)
sin²x+cosx-1=0
1-cos²x+cosx-1=0
-cos²x+cosx=0
cos²x-cosx=0
cosx(cosx-1)=0
cosx=0 cosx-1=0
x=π/2 + πk, k∈Z cosx=1
x=2πk, k∈Z
Ответ: π/2 + πk, k∈Z;
2πk, k∈Z.
3)
![(frac{6x^2-5}{2x-1} )' = frac{(6x^2-5)'(2x-1)-(6x^2-5)(2x-1)'}{(2x-1)^2}= \ \
= frac{12x(2x-1)-2(6x^2-5)}{(2x-1)^2}= frac{24x^2-12x-12x^2+10}{(2x-1)^2}= \ \
= frac{12x^2-12x+10}{(2x-1)^2}](https://tex.z-dn.net/?f=+%28frac%7B6x%5E2-5%7D%7B2x-1%7D+%29%27+%3D+frac%7B%286x%5E2-5%29%27%282x-1%29-%286x%5E2-5%29%282x-1%29%27%7D%7B%282x-1%29%5E2%7D%3D+++%5C++%5C+%0A%3D+frac%7B12x%282x-1%29-2%286x%5E2-5%29%7D%7B%282x-1%29%5E2%7D%3D+frac%7B24x%5E2-12x-12x%5E2%2B10%7D%7B%282x-1%29%5E2%7D%3D+%5C++%5C+%0A%3D+frac%7B12x%5E2-12x%2B10%7D%7B%282x-1%29%5E2%7D+++ "(frac{6x^2-5}{2x-1} )' = frac{(6x^2-5)'(2x-1)-(6x^2-5)(2x-1)'}{(2x-1)^2}= \ \
= frac{12x(2x-1)-2(6x^2-5)}{(2x-1)^2}= frac{24x^2-12x-12x^2+10}{(2x-1)^2}= \ \
= frac{12x^2-12x+10}{(2x-1)^2}")
sin²x+cosx-1=0
1-cos²x+cosx-1=0
-cos²x+cosx=0
cos²x-cosx=0
cosx(cosx-1)=0
cosx=0 cosx-1=0
x=π/2 + πk, k∈Z cosx=1
x=2πk, k∈Z
Ответ: π/2 + πk, k∈Z;
2πk, k∈Z.
3)
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