Delta=1*2(-3)+1*1*2+2(-1)(-3)-
-2*2*2-1(-1)(-3)-1*1(-3)=
=-6+2+6-8-3+3=-6
Delta(x1)=(-2)*2(-3)+7*1*2+5(-1)(-3)-
-5*2*2-7(-1)(-3)-(-2)*1(-3)=
=12+14+15-20-21-6=-6
x1=Delta(x1)/Delta=-6/(-6)=1
Delta(x2)=1*7(-3)+1*5*2+2(-2)(-3)-
-2*7*2-1*5(-3)-1(-2)(-3)=
=-21+10+12-28+15-6=-18
x2=Delta(x2)/Delta=-18/(-6)=3
Delta(x3)=1*2*5+2*7(-1)+1*1(-2)-
-2*2(-2)-1*1*7-1*5(-1)=
=10-14-2+8-7+5=0
x3=Delta(x3)/Delta=0/(-6)=0
Ответ: (1;3;0)
(√5)^(㏒₂3) / (√3)^(㏒₂5) = 5 ^((1/2)*㏒₂3) / 3^((1/2)*㏒₂5) =
= ㏒₂ 5 ^((1/2)*㏒₂3) / ㏒₂ 3^((1/2)*㏒₂5) = (1/2)*㏒₂ 3* ㏒₂ 5 / (1/2)*㏒₂5* ㏒₂ 3 =1
Метод Лягерра, точность <span> 1e-3</span>
<span>x1 ≈ 0.666666552553487
<span>P(x1) ≈ 0 <span>iter = </span>4
</span><span>x2<span> ≈ 0.749999727053003 − i ∙ 0.66143770861557
</span>P(x2) ≈ 0 <span>iter = </span>3
</span><span>x3<span> ≈ 0.750000123124048 + i ∙ 0.661438012106969
</span>P(x3) ≈ 0 <span>iter = </span>3
</span><span>x4<span> ≈ 1.0000003286761
</span>P(x4) ≈ 0 <span>iter = </span>1
</span><span>x5<span> ≈ 1.49999993526003
</span>P(x5) ≈ 0 <span>iter = </span>1</span></span>
а(4а-5)(2а+3)=(4а2-5а)(2а+3)=8а3+12а2-10а2-15а=8а3+12а2-10а2-15а=8а3+2а2-15а