Question

un cateto vale x+2 el otro 2x-1 y la hipotenusa vale 9. ¿cuánto vale x?​

Answer 1

Explicación paso a paso:

Hipotenusa²= (cateto 1)²+(cateto 2)²

${9}^{2} = {(x + 2)}^{2} + (2x - 1) {}^{2}$

Para el desarrollo utilizaremos la teoría del binomio al cuadrado

$(x + y) {}^{2} = {x}^{2} + 2xy + {y}^{2} \\ (x - y) {}^{2} = {x}^{2} - 2xy + {y}^{2}$

DESARROLLO

${9}^{2} = (x + 2) {}^{2} + (2x - 1) {}^{2} \\ 81 =( {x}^{2} + 2(x)(2) + {2}^{2} ) + ( {(2x)}^{2} - 2(x)(1) + {1}^{2} ) \\ 81 = ( {x}^{2} + 4x + 4) + (4 {x}^{2} - 4x + 1) \\ 81 = {x}^{2} + 4x + 4 + 4 {x}^{2} - 4x + 1 \\ 81 = 5 {x}^{2} + 5 \\ 81 - 5 = 5 {x}^{2} \\ 76 = 5 {x}^{2} \\ \frac{76}{5} = {x}^{2} \\ 15.2 = {x}^{2} \\ \sqrt{15.2 } = x \\ 3.89871= x$