Am atașat rezolvarea.
Explicație pas cu pas:
[tex]4( {a}^{2} + {b}^{2} + {c}^{2}) = 4(3b + 5c - a) + 14 \\ [/tex]
[tex]4{a}^{2} + 4a + 4{b}^{2} - 12b + 4{c}^{2} - 20c = 14 \\ [/tex]
[tex]{(2a)}^{2} + 2 \cdot 2a + 1 - 1 + {(2b)}^{2} - 2 \cdot 6b + 9 - 9 + {(2c)}^{2} - 2 \cdot 10c + 25 - 25 = 14 \\ [/tex]
[tex]{(2a + 1)}^{2} + {(2b - 3)}^{2} + {(2c - 5)}^{2} = 14 + 35 \\ [/tex]
[tex]\underbrace{{(2a + 1)}^{2}}_{ \geqslant 0} + \underbrace{{(2b - 3)}^{2}}_{ \geqslant 0} + \underbrace{{(2c - 5)}^{2}}_{ \geqslant 0} = {7}^{2} \\ [/tex]
[tex]- 7 \leqslant 2a + 1 \leqslant 7 \iff - 8 \leqslant 2a \leqslant 6 \\ [/tex]
[tex]- 4 \leqslant a \leqslant 3 \implies \bf a \in \Big[ - 4 ; 3\Big][/tex]
[tex]- 7 \leqslant 2b - 3 \leqslant 7 \iff - 4 \leqslant 2b \leqslant 10 \\ [/tex]
[tex]- 2 \leqslant b \leqslant 5 \implies \bf b \in \Big[ - 2 ; 5\Big][/tex]
[tex]- 7 \leqslant 2c - 5 \leqslant 7 \iff - 2 \leqslant 2c \leqslant 12 \\ [/tex]
[tex]- 1 \leqslant c \leqslant 6 \implies \bf c \in \Big[ - 1 ; 6\Big][/tex]
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Am atașat rezolvarea.
Explicație pas cu pas:
[tex]4( {a}^{2} + {b}^{2} + {c}^{2}) = 4(3b + 5c - a) + 14 \\ [/tex]
[tex]4{a}^{2} + 4a + 4{b}^{2} - 12b + 4{c}^{2} - 20c = 14 \\ [/tex]
[tex]{(2a)}^{2} + 2 \cdot 2a + 1 - 1 + {(2b)}^{2} - 2 \cdot 6b + 9 - 9 + {(2c)}^{2} - 2 \cdot 10c + 25 - 25 = 14 \\ [/tex]
[tex]{(2a + 1)}^{2} + {(2b - 3)}^{2} + {(2c - 5)}^{2} = 14 + 35 \\ [/tex]
[tex]\underbrace{{(2a + 1)}^{2}}_{ \geqslant 0} + \underbrace{{(2b - 3)}^{2}}_{ \geqslant 0} + \underbrace{{(2c - 5)}^{2}}_{ \geqslant 0} = {7}^{2} \\ [/tex]
[tex]- 7 \leqslant 2a + 1 \leqslant 7 \iff - 8 \leqslant 2a \leqslant 6 \\ [/tex]
[tex]- 4 \leqslant a \leqslant 3 \implies \bf a \in \Big[ - 4 ; 3\Big][/tex]
[tex]- 7 \leqslant 2b - 3 \leqslant 7 \iff - 4 \leqslant 2b \leqslant 10 \\ [/tex]
[tex]- 2 \leqslant b \leqslant 5 \implies \bf b \in \Big[ - 2 ; 5\Big][/tex]
[tex]- 7 \leqslant 2c - 5 \leqslant 7 \iff - 2 \leqslant 2c \leqslant 12 \\ [/tex]
[tex]- 1 \leqslant c \leqslant 6 \implies \bf c \in \Big[ - 1 ; 6\Big][/tex]