Question

Решите уравнение пж. Даю спасибо

Answer 1

Ответ:

$1,25 \ ; \ -\dfrac{1}{3} \ ; \ 1 \pm \dfrac{2\sqrt{7}}{7} \ ;$

Объяснение:

$(2x-5)^{2}=4x^{2};$

$(2x)^{2}-2 \cdot 2x \cdot 5+5^{2}=4x^{2};$

$2^{2}x^{2}-20x+25=4x^{2};$

$4x^{2}-4x^{2}-20x+25=0;$

$-20x=-25 \quad | :(-5)$

$4x=5;$

$x=5:4;$

$x=1,25;$

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$(3x+1)^{2}=3x(3x+1);$

$(3x)^{2}+2 \cdot 3x \cdot 1+1^{2}=3x \cdot 3x+3x \cdot 1;$

$3^{2}x^{2}+6x+1=9x^{2}+3x;$

$9x^{2}-9x^{2}+6x-3x=-1;$

$(6-3)x=-1;$

$3x=-1;$

$x=-\dfrac{1}{3};$

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$(7x-5)^{2}=7(x+1)^{2};$

$(7x)^{2}-2 \cdot 7x \cdot 5+5^{2}=7(x^{2}+2 \cdot x \cdot 1+1^{2});$

$7^{2}x^{2}-70x+25=7(x^{2}+2x+1);$

$49x^{2}-70x+25=7x^{2}+14x+7;$

$49x^{2}-7x^{2}-70x-14x+25-7=0;$

$(49-7)x^{2}-(70+14)x+18=0;$

$42x^{2}-84x+18=0 \quad | :6$

$7x^{2}-14x+3=0;$

$D=b^{2}-4ac \Rightarrow D=(-14)^{2}-4 \cdot 7 \cdot 3=196-84=112;$

$x_{1,2}=\dfrac{-b \pm \sqrt{D}}{2a} \Rightarrow x_{1,2}=\dfrac{-(-14) \pm \sqrt{112}}{2 \cdot 7}=\dfrac{14 \pm 4\sqrt{7}}{14}=1 \pm \dfrac{2\sqrt{7}}{7};$