Предмет: Алгебра, автор: grishinalarisa

решите уравнение: корень третьей степени из х-1 + корень третьей степени из 10-х =3

Ответы

Автор ответа: artalex74

$\sqrt[3]{x-1} + \sqrt[3]{10-x}=3 \\ \sqrt[3]{10-x}=3-\sqrt[3]{x-1} \\ (\sqrt[3]{10-x})^3=(3-\sqrt[3]{x-1})^3 \\ 10-x=27-27\sqrt[3]{x-1}+9(\sqrt[3]{x-1})^2-x+1 \\ 9(\sqrt[3]{x-1})^2-27\sqrt[3]{x-1}+18=0 \\ \\ \sqrt[3]{x-1}=t \\ \\ 9t^2-27t+18=0 \\ t^2-3t+2=0 \\ t_1=1,t_2=2$
$\sqrt[3]{x-1}=1$ или $\sqrt[3]{x-1}=2$

х - 1 = 1 или х - 1 = 8
х = 2 или = 9
Ответ: 2; 9.

Автор ответа: mukus13

Воспользуемся формулой: $(a+b)^3=a^3+b^3+3ab(a+b)$

$\sqrt[3]{x-1} + \sqrt[3]{10-x} =3$
$( \sqrt[3]{x-1} + \sqrt[3]{10-x})^3 =3^3$
$x-1+10-x+3 \sqrt[3]{(x-1)(10-x)} *( \sqrt[3]{x-1} + \sqrt[3]{10-x})=27$
учитывая, что $\sqrt[3]{x-1} + \sqrt[3]{10-x} =3$ , то $x-1+10-x+3 \sqrt[3]{(x-1)(10-x)} *3=27$
$9 \sqrt[3]{(x-1)(10-x)}=18$
$\sqrt[3]{(x-1)(10-x)}=2$
$( \sqrt[3]{(x-1)(10-x)})^3=2^3$
$(x-1)(10-x)}=8$
$- x^{2} +10x+x-10-8=0$
$- x^{2} +11x-18=0$
$x^{2} -11x+18=0$
$D=121-72=49$
$x_1=2$
$x_2=9$
Ответ: 2; 9

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