A)2√2/√2√2=2√2/2=√2
б)6√3/√3√3=6√3/3=2√3
в)√(x-y)/√(x-y)√(x-y)= √(x-y)/(x-y)
г)(а+b)√(a+b)/√(a+b)√(a+b)=(a+b)√(a+b)/(a+b)= √(a+b)
д)(х+3)√(х²-9)/√(х²-9)√(х²-9)=(х+3)√(х²-9)/(х²-9)=(х+3)√(х²-9)/(х+3)(х-3)= √(х²-9)/(х-3)
е)(а-b)√(a²-b²)/√(a²-b²)√(a²-b²)=(a-b)√(a²-b²)/(a²-b²)=(a-b)√(a²-b²)/(a-b)(a+b)= √(a²-b²)/(a+b)
ж)(1-√2)/(1+√2)(1-√2)= (1-√2)/(1-2)=- (1-√2)=√2-1
з) (1+√2)/(1+√2)(1-√2)= (1+√2)/(1-2)= -(1+√2)=-1-√2
и)√3(√3+5)/(√3+5)(√3-5)=√3(√3+5)/(3-25)= √3(√3+5)/(-22)=-√3(√3+5)/22
к)а(√а+а)/(√а-а)(√а+а)=а(√а+а)/(а-а²)
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>
Или (7/8-4)×8
(-25/8)×8
-25
Ответ:8n+4m/(n+2m)²=12mn/4m²n²=3/mn
Объяснение: