Решите уравнения 1)√x=3 2)√x-3=2 3)⁴√x-3=2 4)³√x+2=3 5)√x+3=-2 6)³√x+3=2 7)√x=2-x 8)√x-2=x/3. Created by Куроко68. matematika-ru

Ответ: , так как , то x∈∅;

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Куроко68 @Куроко68

July 2022 1 7 Report

Решите уравнения
1)√x=3
2)√x-3=2
3)⁴√x-3=2
4)³√x+2=3
5)√x+3=-2
6)³√x+3=2
7)√x=2-x
8)√x-2=x/3

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Ответ:

$\tt \displaystyle 1) \; \; \sqrt{x} =3 \Leftrightarrow \left \{ {{x\geq 0} \atop {(\sqrt{x})^2 =3^2}} \right. \Leftrightarrow \left \{ {{x\geq 0} \atop {x = 9}} \right. \Rightarrow x=9;$

$\tt \displaystyle 2) \; \; \sqrt{x-3} =2 \Leftrightarrow \left \{ {{x-3\geq 0} \atop {(\sqrt{x-3})^2 =2^2}} \right. \Leftrightarrow \left \{ {{x\geq 3} \atop {x-3 = 4}} \right. \Leftrightarrow \left \{ {{x\geq 3} \atop {x = 7}} \right. \Rightarrow x=7;$

$\tt \displaystyle 3) \; \; \sqrt[4]{x-3} =2 \Leftrightarrow \left \{ {{x-3\geq 0} \atop {(\sqrt{x-3})^4 =2^4}} \right. \Leftrightarrow \left \{ {{x\geq 3} \atop {x-3 = 16}} \right. \Leftrightarrow \left \{ {{x\geq 3} \atop {x = 19}} \right. \Rightarrow x=19;$

$\tt \displaystyle 4) \; \; \sqrt[3]{x+2} =3 \Leftrightarrow (\sqrt{x+2})^3 =3^3 \Leftrightarrow x+2=27 \Leftrightarrow x=25;$

$\tt \displaystyle 5) \; \; \sqrt{x+3} =-2$ , так как $\tt \displaystyle \sqrt{x+3} \geq 0, -2$ , то x∈∅;

$\tt \displaystyle 6) \; \; \sqrt[3]{x+3} =2 \Leftrightarrow (\sqrt{x+3})^3 =2^3 \Leftrightarrow x+3=8 \Leftrightarrow x=5;$

$\tt \displaystyle 7) \; \; \sqrt{x} =2-x \Leftrightarrow \left \{ {{x\geq 0, \; 2-x\geq 0} \atop {(\sqrt{x})^2 =(2-x)^2}} \right. \Leftrightarrow \left \{ {{x\geq 0, \; 2\geq x} \atop {x = 4-4 \cdot x+x^2}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{0 \leq x \leq 2} \atop {x^2-5 \cdot x+4=0}} \right. \Leftrightarrow \left \{ {{0 \leq x \leq 2} \atop {(x-1) \cdot (x-4)=0}} \right. \Leftrightarrow \left \{ {{0 \leq x \leq 2} \atop {x_{1}=1, \; x_{2}= 4}} \right. \Rightarrow \\\\ \Rightarrow x=1;$

$\tt \displaystyle 8) \; \; \sqrt{x-2} =\frac{x}{3} \Leftrightarrow \left \{ {{x-2\geq 0,\; \dfrac{x}{3} \geq 0} \atop {(\sqrt{x-2})^2 =(\dfrac{x}{3})^2}} \right. \Leftrightarrow \left \{ {{x\geq 2,\; x \geq 0} \atop {x-3 = 4}} \right. \Leftrightarrow \left \{ {{x\geq 2} \atop {x-2 = \dfrac{x^2}{9}}} \right. \Leftrightarrow \\\\ \Leftrightarrow \left \{ {{x\geq 2} \atop {x^2-9 \cdot x+18=0}} \right. \Leftrightarrow \left \{ {{x\geq 2} \atop {(x-3) \cdot (x-6)=0}} \right. \Leftrightarrow$

$\tt \displaystyle \Leftrightarrow \left \{ {{x\geq 2} \atop {x_{1}=3,\; x_{2}=6}} \right. \Rightarrow x \in \{3; \; 6 \}.$

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