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Bună,
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[tex]2(x + 15) ^{2} = 32 | \div 2 \\ (x + 15) ^{2} = 16 \\ x + 15 = ±4 \\ x_1 + 15 = 4 = > \green{ x_1 = - 11} \\ x_2 + 15= - 4 = > \pink{ x_2 = - 19}[/tex]
[tex]b)3(x - 12) ^{2} = 75 \: | \div 3 \\( x - 12) ^{2} = 25 \\ x - 12 = ±5 \\ x_1 - 12 = 5 = > \blue{ x_1 = 17 }\\ x_2 - 12 = - 5 = > \purple{x_2 = 7}[/tex]
[tex](4x - 5 \sqrt{5} ) ^{2} = 45 \\ 4x - 5 \sqrt{5} = ±3 \sqrt{5} = > \\ 4x_1 - 5 \sqrt{5} = 3 \sqrt{5} \\ 4x_1 = 3 \sqrt{5} + 5 \sqrt{5} \\ 4x_1 = 8 \sqrt{5} \\ = \orange{ x _1 = 2 \sqrt{5}} \\ = > 4x _2 - 5 \sqrt{5} = - 3 \sqrt{5} \\ 4x_2 = - 3 \sqrt{5} + 5 \sqrt{5} \\ 4x_2 = 2 \sqrt{5} \\ = \red{ x_2 = \frac{ \sqrt{5} }{2}} [/tex]
[tex](2 \sqrt{6} + 3x) ^{2} = 96 \\ 2 \sqrt{6} + 3x = ±4 \sqrt{6} \\ 2 \sqrt{6} + 3x_1= 4 \sqrt{6} \\ 3x_1 = 4 \sqrt{6} - 2 \sqrt{6} \\ 3x_1= 2 \sqrt{6} \\ \blue{ x _1= \frac{2 \sqrt{6} }{3} } \\ 2 \sqrt{6} + 3x_2 = - 4 \sqrt{6} \\ 3x_2 = - 4 \sqrt{6} - 2 \sqrt{3} \\ 3x_2 = - 6 \sqrt{3} \\ \green{ x_2 = - 2 \sqrt{6} }[/tex]
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Sper că te-am ajutat.
