Алгебра: решить задание 50,51,51,53,54по Алгебре 10класс

решить задание 50,51,51,53,54
по Алгебре 10класс

решить задание 50,51,51,53,54по Алгебре 10класс

Ответ:

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28.11.2020 19:57

$50)\ \ \sqrt{7-x}=x-1\ \ \ \Leftrightarrow \ \ \ \ \left\{\begin{array}{l}x-1\geq 0\\7-x=(x-1)^2\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1\\7-x=x^2-2x+1\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq 1\\x^2-x-6=0\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1\\x_1=-2\ ,\ x_2=3\end{array}\right\ \ \ \Rightarrow \ \ \underline {\ x=3\in [\, 2\, ;\, 4\, ]\ }$

$51)\ \ \sqrt{3-x}=x-1\ \ \ \Leftrightarrow \ \ \ \ \left\{\begin{array}{l}x-1\geq 0\\3-x=(x-1)^2\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1\\3-x=x^2-2x+1\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq 1\\x^2-x-2=0\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1\\x_1=-1\ ,\ x_2=2\end{array}\right\ \ \ \Rightarrow \ \ \underline {\ x=2\ }$

$53)\ \ \sqrt{2-x}=x\ \ \ \Leftrightarrow \ \ \ \ \left\{\begin{array}{l}x\geq 0\\2-x=x^2\end{array}\right\ \ \left\{\begin{array}{l}x\geq 0\\x^2+x-2=0\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq 0\\x_1=-2\ ,\ x_2=1\end{array}\right\ \ \ \Rightarrow \ \ \underline {\ x=1\ }$

$54)\ \ \sqrt{3x-5}=-x+5\ \ \ \Leftrightarrow \ \ \ \left\{\begin{array}{l}-x+5\geq 0\\3x-5=(5-x)^2\end{array}\right\ \ \left\{\begin{array}{l}5\geq x\\3x-5=25-10x+x^2\end{array}\right\\\\\\\left\{\begin{array}{l}x\leq 5\\x^2-13x+30=0\end{array}\right\ \ \left\{\begin{array}{l}x\leq 5\\x_1=10\ ,\ x_2=3\end{array}\right\ \ \ \Rightarrow \ \ \underline {\ x=3\in [\, 1,5\, ;\, 3,5\, ] }$