<span><span>(<span><span>a2</span>−<span>b2</span>+ab−2</span>)</span><span>(<span><span>a2</span>−<span>b2</span>−ab−2</span>)</span>−<span><span>(<span><span>a2</span>−<span>b2</span>−2</span>)</span>2</span>=</span>Возведение в степень:<span><span>(<span><span>a2</span>−<span>b2</span>+ab−2</span>)</span><span>(<span><span>a2</span>−<span>b2</span>−ab−2</span>)</span>−<span>(<span><span>a4</span>−2<span>a2</span><span>b2</span>−4<span>a2</span>+<span>b4</span>+4<span>b2</span>+4</span>)</span>=</span>Раскрытие скобок:<span><span><span>a4</span>+<span>a2</span><span>(<span>−1</span>)</span><span>b2</span>+<span>a3</span><span>(<span>−1</span>)</span>b+<span>a2</span><span>(<span>−2</span>)</span>−<span>b2</span><span>a2</span>+<span>b4</span>+<span>b3</span>a+2<span>b2</span>+<span>a3</span>b+a<span>b3</span><span>(<span>−1</span>)</span></span><span>+<span>a2</span><span>b2</span><span>(<span>−1</span>)</span>+ab<span>(<span>−2</span>)</span>−2<span>a2</span>+2<span>b2</span>+2ab+4−<span>a4</span>+2<span>a2</span><span>b2</span>+4<span>a2</span>−<span>b4</span>−4<span>b2</span>−4=</span></span><span><span>a2</span><span>(<span>−1</span>)</span><span>b2</span>−<span>b2</span><span>a2</span>+<span>b3</span>a+a<span>b3</span><span>(<span>−1</span>)</span>+<span>a2</span><span>b2</span>=</span><span>−<span>a2</span><span>b2</span>−<span>b2</span><span>a2</span>+<span>b3</span>a−a<span>b3</span>+<span>a2</span><span>b2</span>=</span>−a2b2
A²(a+1)-4 (a+1)= а³+а²-4а-4= а5- 4а-4
Удачи :)
Чуть больше 2, но чуть меньше 3
Из этой формулы можно найти b1.
Теперь вычислим сумму первых восьми членов прогрессии.
<ADM и <BCN равны, потому что оба равнобедренные и параллельны друг другу, или же симметричны.