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>
<span>(2а)^3 × 4а^-2 / (4а^3)^2=
= 8a^3 * 4a^-2 / 8a^5 =
= 4a / a^5 =
= 4 / a^4
</span>
Надеюсь я правильно поняла
(2c^3−7d^2)⋅(2c^3+7d^2)=2c^3 - 7d^2
3х в квадрате это =9х
9х-13х+14=0
-4х+14=0
-4х=-14
Х=14:4