<span>-x^2-10x-23=0 |* (-1)
x^2 +10x + 23 =0
D= b^2 - 4ac
D= 100 - 4*1*23 = 8
x1 = -10+ 2</span>√2 / 2 = -5+√2
x2= -5- √2
2 - 5Cosx - Cos2x = 0
2 - 5Cosx - (2Cos²x - 1) = 0
2 - 5Cosx - 2Cos²x + 1 = 0
2Cos²x + 5Cosx - 3 = 0
Cosx = 1/2 Cosx = - 3 - решений нет, так как |- 3| > 1
x = + - π/3 + 2πn, n ∈ z
F(x)=3x-4x²+2
a=45 град
а=tga=tg45=1 a=f`(xo)=1
1)f`(x)=(3x-4x²+2)`=3-4*2x=3-8x
2)3-8xo=1
-8xo=-2
xo=1/4
f(xo)=f(1/4)=3*1/4 - 4*(1/4)²+2=3/4 -1/4 +2=2/4 +2=1/2 +2=2 1/2 =5/2
Ответ: г) (1\4; 5\2)
1)
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25(sin²x<span>+100cosx =89 ;</span>
25(1 -cos²x) +100cosx =89 * * * sin²x =1 -cos²x * * *
25cos²x -100cosx +64 =0 * * * можно замену t = cosx , | t | ≤ 1 * * *
cosx =(50 +√(50² -25*64)/ 25 =(50 +30)/ 25 =80 /25 =16/5 > 1 не годится
cosx =(50 -30)/ 25 = 4 / 5 ;
x = ± arc cos(4 / 5) +2π*n ,n ∈Z.
2)
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tq²x -2tqx =0 ;
tqx(tqx -2) =0 ⇒ [ tqx =0 ; tqx =2 . ⇔ [ x =π*n <span> ; x =arc tq2 + </span><span>π*n , n ∈Z.
3)
</span><span>---</span><span>
cos4x =cos6x</span>⇔ cos6x - cos4x =0 ⇔ -2sin(6x - 4x)/2 * <span>sin(6x+4x)/2 </span>=0 <span>⇔
</span>sinx*sin5x =0⇒[ x = π*n ; 5x = π*n ,n ∈Z.⇔[ x = π*n ; x = (π/5)*n <span>,n ∈Z .
</span>
ответ: (π/5)*k , k ∈Z.
Вот просто паралельно переносишь параболу на четыре клетке влево