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>
A²-6ab+9b²=a²-2*a*3b+(3b)²=(a-3b)²
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1)5x(x-2) - 2x + 4 -4x(x-2)=0
5×(x-2) -2(x-2) -4x(x-2)=0
(x-2)×(5x-2-4x)=0
(x-2)×(x-2)=0
x-2=0
x=2
2)9x-3-2(3x-1)+5x(3x-1)=0
3(3x-1)-2(3x-1)+5x(3x-1)=0
(3x-1)×(3-2-5x)=0
(3x-1)×(1-5x)=0
3x-1=0
x=1/3
1-5x=0
x= -1/5