1) 1,5x^3-2,4y=-1,5-4,8= -6,3
2) -4x^3(x^2-3x+2)=-4x^5+12x^4-4x^3
3) (1-x)(2y+x)=2y+x-2xy-x^2
4) (5c-4)^2=25c^2 - 40c + 16
5) 3a(a-b)+b(2a-b)=3a^2-3ab+2ab-b^2=3a^2-ab-b^2
6) 3c(c-2)-(c-3)^2=3c^2-6c-c^2+6c-9=c^2-9=(c-3)(c+3)
√3sinx +cosx +2cos3x=0 , x∈[π ;3π/2]
2cos(x -π/3) +2cos3x =0 ;
cos3x+<span>cos(x -π/3) =0 ;
2cos(2x - </span>π/6)*cos(x +π/6) =0 ⇔[cos(2x - π/6)=0 ; cos(x +<span>π/6) =0.
</span>* * * cos(2x - π/6)=0 или cos(x +π/6) =0 * * *
[2x - π/6=π/2+π*n ; x +π/6 = π/2+π*n , n∈Z.
[x = π/3+<span>π*n/2 </span> ; x =π/3+π*n , n∈Z .
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x =π/3+π*n/2 ,n∈Z . ⇒x =π/3+π ∈[π ;3π/2] , если n =2 .<span>
x =</span>π/3+π*n , n∈Z . ⇒ x =π/3+π ∈[π ;3π/2] , если n =1 .
ответ: 4π/3.
* * *P.S. a*sinx +b*cosx =√(a²+b²) cos(x -ω) , где ctqω = b/a * * *
√3sinx +cosx =2*((1/2)*cosx +(√3/2)*sinx) =
2*(cosx*cosπ/3 +sinx*sinπ/3) = 2cos(x -π/3 )<span> .
</span>-------
π ≤ π/3+π*n/2 ≤ 3π/2⇔π - π/3 ≤ π*n/2 ≤ 3π/2 -π/3⇔
2π/3 ≤ π*n/2 ≤ 7π/6⇔ 4/3 ≤ n <span>≤ </span>7/3⇒ n=2.
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π ≤ π/3+π*n ≤ 3π/2⇔π - π/3≤ π*n ≤ 3π/2 -π/3⇔2π/3 ≤ π*n ≤ 4π/3<span>⇔
</span>2/3 ≤ n 4/3⇒ n=1
смотри прикреплённый файл
D=b^2-4*a*c
D1=25-4*3*2=1
D2=16-4*4*1=0
D3=4+12=16
D4=9+24=33
3(х-15)=2(х-3)
3х-45=2х-6
х=39
