Question

Решите уравнения. (иррациональные уравнения).

Answer 1

$1)\quad \sqrt{x+6}+\sqrt{x+1}=\sqrt{7x+4}\; ,\\\\ODZ:\; \left \{ {{x+6 \geq 0\; ,\; x+1 \geq 0} \atop {7x+4 \geq 0}} \right. \; , \left \{ {{x \geq -6,\; x \geq -1} \atop {x \geq -4/7}} \right. \; \to \; \; x \geq -\frac{4}{7}\\\\(x+6)+2\sqrt{(x+6)(x+1)}+(x+1)=7x+4\\\\2\sqrt{x^2+7x+6}=5x-3\\\\4(x^2+7x+6)=25x^2-30x+9\\\\21x^2-58x-15=0\\\\D/4=29^2+21\cdot 15=1156\; ,\; \sqrt{D}=34$
$x_1= \frac{29-34}{21} =-\frac{5}{21}\in ODZ\\\\x_2= \frac{29+34}{21}=3\in ODZ$

$Proverka:\; x=3:\; \sqrt{3+6}+\sqrt{3+1}=\sqrt{21+4}\; ;\\\\3+2=5\; ;\\\\5=5.\\\\x=-\frac{5}{21}:\; \; \sqrt{\frac{121}{21}}+\sqrt{\frac{16}{21}}\ne \sqrt{\frac{49}{21}}\; ;\\\\\frac{15}{\sqrt{21}}\ne\frac{7}{\sqrt{21}}\; \; \to \\\\Otvet:\; \; 3\; .$

$2)\quad \sqrt[3]{x+2}-\sqrt[3]{x+17}=1\\\\\star \; \; \underline {(a-b)}^3=a^3-3a^2b+3ab^2-b^3=(a^3-b^3)-3ab\underline {(a-b)}\; \; \star \\\\(x+2)-(x+17)-3\cdot \sqrt[3]{(x+2)(x+17)}\cdot \underbrace {(\sqrt[3]{x+2}-\sqrt[3]{x+17})}_{1}=1^3\\\\19-3\sqrt[3]{x^2+19x+34}=1\\\\\sqrt[3]{x^2+19x+34}=6\\\\x^2+19x+34=6^3\\\\x^2+19x-182=0\\\\D=19^2+4\cdot 182=1089\; ,\; \; \sqrt{D}=33\\\\x_1= \frac{-19-33}{2}=-26\; ,\\\\x_2=\frac{-19+33}{2}=7$

Проверка :   $x=-26:\; \; \sqrt[3]{-24}-\sqrt[3]{-9}\ne 1\; ;$

$-2\sqrt[3]{3}+\sqrt[3]{3^2}\ne 1\\\\x=7:\; \; \sqrt[3]{9}-\sqrt[3]{24}\ne 1\\\\\sqrt[3]3^2-2\sqrt[3]{3}\ne 1\\\\Otvet:\; net\; reshenij\; .$