Решение. Способ 1<span><span>f′</span>(x)=<span><span>(<span><span>4⋅x−7 / </span><span>x2</span></span>)</span>′</span>=
</span><span>=(<span><span><span><span>(4⋅x−7)</span>′</span>⋅<span>x2</span>−<span>(4⋅x−7)</span>⋅<span><span>(<span>x2</span>))</span>′ / </span></span><span><span>(<span>x2</span>)</span>2</span></span>=
</span><span>=<span><span><span><span>(4⋅x)</span>′</span>⋅<span>x2</span>−<span>(4⋅x−7)</span>⋅2⋅x</span><span><span>(<span>x2</span>)</span>2</span></span>=
</span><span>=<span><span>4⋅<span>x2</span>−<span>(4⋅x−7)</span>⋅2⋅x / </span><span><span>(<span>x2</span>)</span>2
</span></span></span>Ответ:<span>f′</span>(x)=<span><span>4⋅<span>x2</span>−<span>(4⋅x−7)</span>⋅2⋅x / </span><span><span>(<span>x2</span>)</span><span>2</span></span></span>
3*(-8)+2у+30=О
-24+2у+30=0
6+2у=0
2у=-6
у=-3
<span>
Решение:Пусть х т - </span>на первом току.
Тогда на втором -
т.
На обоих токах -
т.
Составим и решим уравнение: 
Ответ: 400 т; 600 т
1) (3-а)(9+3а+а2)
2) (4-m)(16+4m+m2)
3) (0,2+a)(0,04-0,2a+a2)
4) (0,3+n)(0,09-0,3n+n2)
2 - степень
6x+3=8x-6y+12
6x-8x=-6y+12-3
-2x=-6y+9
X=3y-9/2
