C ocju x: y=0, xˇ2-x-2=0 , (x-2)(x+1)=0, x1=2, x2=-1
y=xˇ2-x-2, y´=2x-1,y´=k=tg a
a)y´(2)=2.2-1=3, y(2)=0 , P/2,0/
y-y1=k(x-x1)
y-0=3(x-2), y=3x-3, 3x-y-3=0
b)y´(-1)=2.(-1)-1=-3, y(-1)=0 , P´/-1,0/
y-0=-3(x+1), y=-3x-3, 3x+y+3=0
<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
Решение на фото.
Ответ в номере 6 : (-3/4;4) (1;-3) *не уместилось
изначально берем отрицательные значения , дабы в результате игрик вышел не очень большим, а потом наносим на систему координат