X^2-x-6/x-3
Решаем уравнение из числителя
a=1 b=-1 c =-6
D=b^2-4ac=1+24=25
x1=3
x2=-2
Раскладываем по данной формуле:
ax^2+bx+c=a(x-x1)(x-x2)
x^2-x-6=(x-3)(x+2)
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Продолжаем.
(x-3)(x+2)/x-3
x-3 сокращаются
Получаем в ответе x+2
(V7+V3)^2-2V21=(V7)^2+2•V7•V3+(V3)^2-2V21=7+2V21+3-2V21=10
2)(a-1/a)•1/(a-1)=((a^2-1)/a)•1/(a-1)=(a-1)(a+1)/(a•(a-1))=
(a+1)/a
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A)
sin2x +2sinx =√3cosx+√3 ;
<span>2sinx*cosx +2sinx =√3cosx+√3 ;
</span>2sinx(cosx +1) - √3(<span>cosx +1) =0 ;
</span>2(cosx +1) (sinx -<span>√3 /2 ) =0 ;
</span>[ cosx = -1 ; sinx = √3 /2⇔ [ x = (1+2n)π ; x = (-1)^n*π/3 + <span>πn ; n</span>∈Z .
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б)
2 cos²(3π/2+x) =sin2x ; * * * cos(3π/2+x) =sinx * * *
2sin²x<span> =sin2x ;
2sinx(sinx -cosx) =0 ;
[ sinx =0 ; </span>sinx =cosx⇔ [ x =πn ; tqx =1 .⇔ [<span> x =πn </span> ; x =π/4 + πn <span>; n</span>∈Z .
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в)
(sin2x+cosx)(√3+<span>√3*√tgx)=0 ;
</span>2cosx(sinx+1/2)*√3(1 + √tgx ) =0 ; ОДЗ :tgx ≥0
2√3 *cosx(sinx+1/2)*(1 + √tgx ) =0 ;
1 + √tgx ≥ 1 ≠ 0<span>
cosx </span>≠0 не определена <span>tgx ;
</span>{ sinx= -1/2 ; tgx ≥0 . ⇒ x =4π/3+ 2πn , ; n<span>∈Z .
</span> ---------------------------------
<span>г)
</span><span>10*5^(2x-1)-19*35^x+1470*7^(2x-2)=0 ;
</span>10(5^x)²* 5⁻¹ - 19*5^x*7^x + 1470*(7^x)²* 7⁻² =0 ;
2*(5^x)² - 19*5^x*7^x + 30*(7^x)² =0 ;
2*( (5/7)^x)² -19*(5/7)^x +30 =0 ; замена : t = (5/7)^x >0
2t² - <span>19t +30 =0 ;
</span>t₁ =(19-11) /4 = 2 ⇔(5/7) ^x =2 ⇒x₁ =Lq2 /(Lq5 -Lq7).
t ₂ =(19+11) /4 = 15/2 ⇔(5/7) ^x =<span>15/2</span> ⇒x₂ =(Lq15-Lq2) / (Lq5 -Lq7).<span>
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