[tex] \sqrt{x+\sqrt{x+11}} +\sqrt{x-\sqrt{x+11 }} =4 [/tex] Помогите пожалуйста решить. Created by veffi. algebra-ru

Проверка:-верноОтвет: 5

tex] Помогите пожалуйста решить...

veffi @veffi

August 2022 1 10 Report

[tex] \sqrt{x+\sqrt{x+11}} +\sqrt{x-\sqrt{x+11 }} =4 [/tex]
Помогите пожалуйста решить

Answers & Comments

mishka19

Verified answer

$\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}} =4\\\\ (\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}})^2 =4^2\\ \\ (\sqrt{x+\sqrt{x+11}})^2+2\sqrt{x+\sqrt{x+11}}\sqrt{x-\sqrt{x+11}}+(\sqrt{x-\sqrt{x+11}})^2 =4^2\\ \\ x+\sqrt{x+11}+2\sqrt{(x+\sqrt{x+11})(x-\sqrt{x+11})}+x-\sqrt{x+11}=16\\ \\ 2x+2\sqrt{x^2-(\sqrt{x+11})^2} =16\\ \\ 2\sqrt{x^2-(x+11)} =16-2x\\ \\ \sqrt{x^2-x-11} =8-x\\ \\$

$\left \{ {{8-x\geq0} \atop {(\sqrt{x^2-x-11})^2=(8-x)^2}} \right. \\ \\ \left \{ {{-x\geq-8} \atop {x^2-x-11=8^2-2\cdot8\cdot x+x^2}} \right. \\ \\ \left \{ {{x\leq8} \atop {x^2-x-11=64-16x+x^2}} \right. \\ \\ \left \{ {{x\leq8} \atop {15x=75}} \right. \\ \\ \left \{ {{x\leq8} \atop {x=5}} \right. \\ \\ x=5$

Проверка:

$\sqrt{5+\sqrt{5+11}}+\sqrt{5-\sqrt{5+11}} =\sqrt{5+\sqrt{16}}+\sqrt{5-\sqrt{16}} =\\ =\sqrt{5+4}+\sqrt{5-4}=\sqrt{9}+\sqrt{1}=3+1=4$

$4=4$ -верно

Ответ: 5

2 votes Thanks 1

Answers & Comments

Verified answer

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