А1.
(√(6-t) + √(5-t))² =4²
6-t+2<span>√[(6-t)(5-t)]+5-t=16
2</span>√(30-5t-6t+t²<span>)=2t+16-11
(2</span>√(t²-11t+30))²=(2t+5)²
4(t²-11t+30)=4t²+20t+25
4t²-4t²-44t-20t=25-120
<span>-64t=-95
t=95/64
</span>√(6-t) - √(5-t)=√(6 - ⁹⁵/₆₄) - √(5 - ⁹⁵/₆₄)=√(²⁸⁹/₆₄) - <span>√(</span>²²⁵/₆₄<span>)=
</span><span>= </span>¹⁷/₈ - ¹⁵/₈ = ²/₈ = ¹/₄ = 0.25
<span>Ответ: 1)
А2.
[5(x+2)+1]/[(x+2)-4] - [5(x+6)+1]/[(x+6)-4]=
=(5x+10+1)/(x-2) - (5x+30+1)/(x+2)=
=(5x+11)(x-2) - (5x+31)/(x+2) =
=[(5x+11)(x+2)-(5x+31)(x-2)]/(x</span>² -4<span>)=
=(5x</span>²+11x+10x+22-5x²-31x+10x+62<span>)/(x</span>² -4<span>)=
=84/(x</span>² -4<span>)
Ответ: 1)
</span>
Вот так. Всё написано, область определения значений - х не равен 6 и -6.
№159
1) х1+х2=6 х1=5 3)х1+х2=-20 х1=-1
х2*х2=5 х2=1 х1*х2=19 х2=-19
5) х1+х2=-9 х1=-11 2)х1+х2=-4 х1=-5
х1*х2=-22 х2=2 х1*х2=-5 х2=1
4)х1+х2=-2 х1=-1 6)х1+х2=20 х1=-10
х1*х2=1 х2=-1 х1*х2=-300 х2=30