<span>а1=40,d= - 2
a17=a1+16d=40-32=8</span>
(х+1+16/х) (х-20+16/х) +54=0
Замена х+16/х=у
(у+1)(у-20)+54=0
y^2-19y+34=0; y1=2, y2=17
Обратная замена:
1) х+16/х=2
x^2-2x+16=0; D<0
2) x+16/x=17
x^2-17x+16=0
<span>x=1; x=16.</span>
A) (x-4)(x+5)=x^2-4x+5x-20=x^2+x-20
b) (x-1)(2x+3)=2x^2-2x+3x-3=2x^2+x-3
task/30341621 доказать тождество
(2x-2)/(4x²+4x+4) +3x/(2x³-2)+1/(2x-2) = (x-1)/(x²+x+1) ???
<u>решение</u> (2x-2) /(4x²+4x+4) +3x/(2x³-2) +1/(2x-2) =
2(x-1)/4(x²+x+1) +3x/2(x³-1) +1/2(x-1) =
(x-1)/2(x²+x+1) +3x/2(x-1)(x²+x+1) +1/2(x-1) =
( (x-1)² +3x +(x²+x+1) ) / 2(x-1)(x²+x+1) =
(x²-<u>2x</u>+1 +<u>3x</u> +x²+<u>x</u>+1) / 2(x-1)(x²+x+1) =(2x²+2x+2) / 2(x-1)(x²+x+1) =
2(x²+x+1) / 2(x-1)(x²+x+1) = 1/(x-1) .
√72-√288 sin в квадрате 15П/8√72 -√(288) *sin²15π/8 = √72(1 - 2sin²(2π -π/8)) =6√2(1 -2sin²π/8) =
(6√2)*cos2*π/8= (6√2)*cosπ/4 =(6√2)*(1/√2) = 6.