3) (11 + 25x)x + 7 (5x - 7)2 - 3x; 4) 4 + (6 - 11x)2 - вопрос №3112575723446112251216712406 от milashkamilana2005 31.12.2022 00:54

Question

Алгебра: 3) (11 + 25x)x + 7 (5x - 7)2 - 3x; 4) 4 + (6 - 11x)2 25x + x(121x + 3).35.21. 1) 13 + xº(x - 9) (x - 3)3 + 11; 2) 26+ (2 + x) x2 (6 + x);3) 3x - x?(15 + x) -(x + 5)2 - 4x; 4) (4 + x) - 6x x?(x+12) + 1.​

milashkamilana2005 · Answer

1) 3sinx+4cosx=26sinx/2cosx/2+4cos²x/2-4sin²x/2-2sin²x/2-2cos²x/2=06sin²x/2-6sinx/2cosx/2-2cos²x/2=0/2cos²x/23tg²x/2-3tgx/2-1=0tgx/2=a3a²-3a-1=0D=9+12=21a1=(3-√21)/6⇒tgx/2=(3-√21)/6⇒x/2=arctg(3-√21)/2+πn⇒x=2arctg(3-√21)/2+2πn,n∈za2=(3+√21)/6⇒tgx/2=(3+√21)/6⇒x/2=arctg(3+√21)/2+πn⇒x=2arctg(3+√21)/2+2πn,n∈z2) 2 sin² x/4 + 5 cos x/2 = 4 2(1-cosx/2)/2+5 cos x/2 = 4 1-cosx/2+5 cos x/2 = 4 4cosx/2=3cosx/2=3/4x/2=+-arccos0,75+2πn,π∈zx=+-2arccos0,75+4πn,π∈z3) 5 - 4 cos² 3x = 4sin3x5-4+4sin²3x-4sin3x=04sin²3x-4sin3x+1=0(2sin3x-1)²=02sin3x=1sin3x=1/23x=(-1)^n*π/6+πn,n∈zx=(-1)^n*π/18+πn/3,n∈z...

3) (11 + 25x)x + 7 < (5x - 7)2 - 3x; 4) 4 + (6 - 11x)2 > 25x + x(121x + 3).35.21. 1) 13 + xº(x - 9)>(x - 3)3 + 11; 2) 26+ (2 + x) < x2 (6 + x);3) 3x - x?(15 + x) > -(x + 5)2 - 4x; 4) (4 + x) - 6x > x?(x+12) + 1.​

3) (11 + 25x)x + 7 < (5x - 7)2 - 3x; 4) 4 + (6 - 11x)2 > 25x + x(121x + 3).
35.21. 1) 13 + xº(x - 9)>(x - 3)3 + 11; 2) 26+ (2 + x) < x2 (6 + x);
3) 3x - x?(15 + x) > -(x + 5)2 - 4x; 4) (4 + x) - 6x > x?(x+12) + 1.