№ 1. Решите неравенства: 1) 3(х-2) х-12; 2) 1,4х-3≥2(0,5х-2,6); 3) 4(1-х)-3(х+2) 5; 4) (х+3)(х-3) ≤ - id1133689520220805 от kvas1985 05.08.2022 10:25

Question

Алгебра: № 1. Решите неравенства: 1) 3(х-2) х-12; 2) 1,4х-3≥2(0,5х-2,6); 3) 4(1-х)-3(х+2) 5; 4) (х+3)(х-3) ≤ (х+3) ^2;

kvas1985 · Accepted Answer

1)2*1/2(sin8x+sin4x)=sin8x+1
sin8x+sin4x=sin8x+1
sin4x=1
4x=π/2+2πn
x=π/8+πn/2,n∈Z
3)sin7x-sinx+cos²2x-sin²2x=0
2sin3xcos4x+cos4x=0
cos4x(2sin3x+1)=0
cos4x=0
4x=π/2+πn
x=π/8+πn/4,n∈z
sin3x=-1/2
3x=(-1)^n+1*π/6+πn
x=(-1)^n+1*π/18+πn,n∈z
2)-4sinxsin5x=2-2cos²2x
-4*1/2(cos4x-cos6x)+cos4x=1
2cos6x-2cos4x+cos4x=1
2cos6x-cos4x=1
2cos6x=cos4x+1
2cos4xcos2x-2sin4xsin2x=2cos²2x
2cos2x(2cos²2x-1)-4cos2xsin²2x-2cos²2x=0
4cos³2x-2cos2x-4cos2x(1-cos²2x)-2cos²2x=0
4cos³2x-2cos2x-4cos2x+4cos³2x-2cos²2x=0
8cos³2x-2cos²2x-6cos2x=0
2cos2x(4cos²2x-cos2x-3)=0
cos2x=0
2x=π/2+πn
x=π/4+πn/2,n∈z
4cos²2x-cosx-3=0
cos2x=a
4a²-a-3=0
D=1+48=49
a1=(1-7)/8=-3/4
cos2x=-3/4
2x=+-(π-arccos3/4)+2πn
x=+-1/2(π-arccos3/4)+πn
a2=(1+7)/8=1
cos2x=2πn
x=πn,n∈z

№ 1. Решите неравенства:
1) 3(х-2)>х-12;
2) 1,4х-3≥2(0,5х-2,6);
3) 4(1-х)-3(х+2)<5;
4) (х+3)(х-3) ≤ (х+3) ^2;

MOGZ ответил

№ 1. Решите неравенства: 1) 3(х-2)>х-12; 2) 1,4х-3≥2(0,5х-2,6); 3) 4(1-х)-3(х+2)<5; 4) (х+3)(х-3) ≤ (х+3) ^2;

MOGZ ответил

№ 1. Решите неравенства:
1) 3(х-2)>х-12;
2) 1,4х-3≥2(0,5х-2,6);
3) 4(1-х)-3(х+2)<5;
4) (х+3)(х-3) ≤ (х+3) ^2;