x₁ = 2,6
x₂ = 4
[tex]\frac{2}{(x-3)^2} + \frac{3}{x-3} - 5 = 0 \\[/tex]
[tex](x-3)^2 \neq 0\\x-3 \neq 0\\x \neq 3[/tex]
[tex]\frac{2+3(x-3)-5(x-3)^2}{(x-3)^2} = 0[/tex]
[tex]2+3x-9 - 5(x^2-6x+9) = 0\\2 +3x-9 - 5x^2+30x-45=0[/tex]
[tex]-5x^2+33x-52 = 0[/tex]
[tex]5x^2 - 33x +52 = 0[/tex]
[tex]D = -33^2- 52*4*5 = 1089 - 1040 = 49 = 7^2[/tex]
[tex]x1 = \frac{33 - \sqrt{D} }{5*2} = \frac{33-7}{10} = \frac{26}{10} = 2,6\\x2 = \frac{33 + \sqrt{D} }{5*2} = \frac{33+7}{10} = \frac{40}{10} = 4[/tex]
[tex]\left \{ {{\left \{ {{x1=2,6} \atop {x2=4}} \right. } \atop {x\neq 3}} \right.[/tex]
ОТВЕТ: x₁ = 2,6
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Answers & Comments
Ответ:
x₁ = 2,6
x₂ = 4
Пошаговое объяснение:
[tex]\frac{2}{(x-3)^2} + \frac{3}{x-3} - 5 = 0 \\[/tex]
[tex](x-3)^2 \neq 0\\x-3 \neq 0\\x \neq 3[/tex]
[tex]\frac{2+3(x-3)-5(x-3)^2}{(x-3)^2} = 0[/tex]
[tex]2+3x-9 - 5(x^2-6x+9) = 0\\2 +3x-9 - 5x^2+30x-45=0[/tex]
[tex]-5x^2+33x-52 = 0[/tex]
[tex]5x^2 - 33x +52 = 0[/tex]
[tex]D = -33^2- 52*4*5 = 1089 - 1040 = 49 = 7^2[/tex]
[tex]x1 = \frac{33 - \sqrt{D} }{5*2} = \frac{33-7}{10} = \frac{26}{10} = 2,6\\x2 = \frac{33 + \sqrt{D} }{5*2} = \frac{33+7}{10} = \frac{40}{10} = 4[/tex]
[tex]\left \{ {{\left \{ {{x1=2,6} \atop {x2=4}} \right. } \atop {x\neq 3}} \right.[/tex]
ОТВЕТ: x₁ = 2,6
x₂ = 4