Question

X+11=10ײ como resolver con el TCP

Answer 1

Respuesta:

$x + 11 = 10x {}^{2}$

$x + 11 - 10x {}^{2} = 0$

$- 10x {}^{2} + x + 11 = 0$

$- 10x {}^{2} + x = - 11$

$- 1( - 10x {}^{2} + x ) = - 1( - 11)$

$10x {}^{2} - x = 11$

$\frac{10x {}^{2} - x }{10} = \frac{11}{10}$

$x {}^{2} - \frac{1}{10}x = \frac{11}{10}$

$x {}^{2} - \frac{1}{10}x + (\frac{ \frac{1}{10} }{2} ) {}^{2} = \frac{11}{10} + ( \frac{ \frac{1}{10} }{2} ) {}^{2}$

$x {}^{2} - \frac{1}{10} x + ( \frac{1}{20} ) {}^{2} = \frac{11}{10} + (\frac{1}{20} ) {}^{2}$

$x {}^{2} - \frac{1}{10} x+ ( \frac{1}{20} ) {}^{2} = \frac{11}{10} + \frac{1 {}^{2} }{20 {}^{2} }$

$x {}^{2} - \frac{1}{10}x + ( \frac{1}{20} ) {}^{2} = \frac{11}{10} + \frac{1}{400}$

$x {}^{2} - \frac{1}{10} x + ( \frac{1}{20} ) {}^{2} = \frac{(400 \div 10)11}{400} + \frac{1}{400}$

$x {}^{2} - \frac{1}{10} x + ( \frac{1}{20} ) {}^{2} = \frac{(40)11}{400} + \frac{1}{400}$

$x {}^{2} - \frac{1}{10} x + ( \frac{1}{20} ) {}^{2} = \frac{440}{400} + \frac{1}{400}$

$x {}^{2} - \frac{1}{10} x + (\frac{1}{20} ) {}^{2} = \frac{440 + 1}{400}$

$(x - \frac{1}{20} ) {}^{2} = \frac{441}{400}$

$x - \frac{1}{20} = + - \sqrt{ \frac{441}{400} }$

$x - \frac{1}{20} = + - \frac{ \sqrt{441} }{ \sqrt{400}}$

$x - \frac{1}{20} = + - \frac{21}{20}$

$x = \frac{1}{20} + - \frac{21}{20}$

$x1 = \frac{1}{20} + \frac{21}{20}$

$x1 = \frac{1 + 21}{20}$

$x1 = \frac{22}{20}$

$x1 = \frac{11}{10}$

$x2 = \frac{1}{20} - \frac{21}{20}$

$x2 = \frac{1 - 21}{20}$

$x2 = \frac{ - 20}{20}$

$x2 = - 1$