2ˣ-3=√(25+9·2ˣ)
одз: 25+9·2ˣ≥0 х∈R т.к. 2ˣ-показательная функция 2>0 2ˣ>0
(2ˣ-3)²=(√(25+9·2ˣ))²
(2ˣ)²-6·2ˣ+9=25+9·2ˣ
(2ˣ)²-15·2ˣ-16=0
2ˣ=t
t²-15t-16=0
t₁+t₂=15
t₁t₂=-16
t₁=-1 2ˣ=-1 нет решения
t₂=16 2ˣ=16
2ˣ=2⁴
x=4
<span>а7=98 а14=42
a7 = a1 + 6b
a14 = a1 13b
a1+6b -98 = 0
a1 +13b -42 = 0
</span>
a1+6b -98 = <span>a1 +13b -42
</span>
6b -98 = <span>13b -42
-98 +42 = 13b - 6b
7b = -56
b = -8
a7 = a1 +6b
98 = a1 -6*8
a1 = 98 +48
a1 = 146
</span>
<span>а6+а14=30
a14 = 30 - a6
a1 +13b = 30 -a1 -5b
2a1 +18b = 30
a1 + 9b = 15
a10 = 15
</span>
( Х - 1 )/5 = ( 5 - Х )/2 + ( 3х/4 )
<span>4( Х - 1 ) = 10( 5 - Х ) + 15х </span>
<span>4х - 4 = 50 - 10х + 15х </span>
<span>4х - 4 = 50 + 5х </span>
<span>5х - 4х = - 4 - 50 </span>
<span>Х = - 54</span>
<span>(c-2)(c+3)-c в квадрате=с^2+3c-2c-6-c^2=c-6</span>
<span>ответ: с-6</span>
3)√6*sin63π/4*tq64π/3 =√6*sin(16π-π/4)*tq(22π-π/3) =-√6*s(-sinπ/4)*(-tqπ/3)=
√6*1/√2/ *√3 =3.
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4) ( 1-2sin²54°)/(8tq9°sin²99°) =cos2*54°/((8tq9°* sin²(90° +9°))
=cos108°/(8tq9°*cos²9°)=cos(90°+18°)/(8sin9°*cos9°) = -sin18°/4sin18° == -1/4.
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5) √2*(sin40°cos5°-sin(180°+50°)sin5°)/(sin25°sin35° -sin(90°+25°)cos35°)=
= -√2*sin45°/*cos60° = -√2*1/√2/1(1/2) = -2.
