Question

Помогите, срочно!!!!

Answer 1

[ log₀,₃(x+1)/(log₀,₃100-log₀,₃9) ] <1
[ log₀,₃(x+1)/(log₀,₃100-log₀,₃9) -1]<0

[ log₀,₃(x+1)-log₀,₃100+log₀,₃9] /(log₀,₃100-log₀,₃9)<0
[ log₀,₃(x+1)+log₀,₃0,09 ] / [-log₀,₃0,09]<0 |*(-1)
[ log₀,₃(x+1)+log₀,₃0,09 ] / 2>0

{log₀,₃(x+1)+log₀,₃0,09>0
 2>0

log₀,₃(x+1)+log₀,₃0,09>0
log₀,₃(x+1)>-log₀,₃0,09
log₀,₃(x+1)>log₀,₃0,09⁻¹       0,09⁻¹=(9/100)⁻¹=100/9
0<0,3<1 знак неравенства меняем

{x+1<(100/9)        {x<(91/9)
x+1>0                    x>-1
x∈(-1;91/9)
наибольшее целое решение х=10

cosα=-(1/3), sinα=?
sin²α+cosα=1
sin²α+(-1/3)²=1, sin²α=8/9
sinα=+-√(8/9),  π<α<3π/2. ⇒
sinα=-√(8/9)
sinα=-2√2/3

sin135°+cos210°-sin(-150°)+cos225°=sin(90°+45°)+cos(180°-30°)+sin(180°-30°)+cos(180°+45°)=cos45-cos30°+sin30°-cos45°=sin30°-cos30°=1/2-√3/2=(1-√3)/2