Question

Решить уравнения:
$\sqrt[3]{1 + \sqrt{ x} } + \sqrt[3]{1 - \sqrt{x} } = 2$
$\sqrt[3]{x + 10} - \sqrt[3]{x - 9} = 1$

Answer 1

$1)\; \; \sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=2\; \; ,\; \; \; ODZ:\; x\geq 0\; ,\\\\\star \; \; (a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)\; \star \\\\( \sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}})^3=2^3\\\\(1+\sqrt{x})+(1-\sqrt{x})+3\cdot \sqrt[3]{(1-\sqrt{x})(1+\sqrt{x})}\cdot \underbrace {(\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}}_{2})=8\\\\2+3\cdot \sqrt[3]{1-x}\cdot 2=8\\\\6\cdot \sqrt[3]{1-x}=6\\\\\sqrt[3]{1-x}=1\\\\(\sqrt[3]{1-x})^3=1^3\\\\1-x=1\\\\\boxed {x=0}$

$2)\; \; \sqrt[3]{x+10}-\sqrt[3]{x-9}=1\\\\\star \; \; (a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab(a-b)\; \; \star \\\\(\sqrt[3]{x+10}-\sqrt[3]{x-9})^3=1^3\\\\x+10-(x-9)-3\sqrt[3]{(x+10)(x-9)}\cdot \underbrace {(\sqrt[3]{x+10}-\sqrt[3]{x-9})}_{1}=1\\\\19-3\cdot \sqrt[3]{x^2+x-90}=1\\\\3\cdot \sqrt[3]{x^2+x-90}=18\\\\\sqrt[3]{x^2+x-90}=6\\\\(\sqrt[3]{x^2+x-90})^3=6^3\\\\x^2+x-90=216\\\\x^2+x-306=0\; \; ,\; \; D=1225\\\\\boxed {\; x_1=-18\; ,\; x_2=17\; }$