Respuestas
Respuesta dada por:
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Como se resuelve: ( 3 raiz de x + 1/2x) a la 2da potencia
![================================== ==================================](https://tex.z-dn.net/?f=%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D)
![Aplicamos\ binomio\ al\ cuadrado: \\ \\\boxed{(x+y)^{2}=x^{2}+2xy+y^{2} }\\ \ \ \\ \\ ===========================\\ \\ \\ \left(3 \sqrt{x}+ \dfrac{x}{2}\right)^{2} = \\ \\ \\ (3 \sqrt{x})^{2}+\not{2}(3 \sqrt{x} )\left( \dfrac{x}{\not{2}} \right) + \left( \dfrac{x}{2} \right)^{2} = \\ \\ \\ \boxed{9x + 3x\sqrt{x} + \dfrac{ x^{2}}{4}} Aplicamos\ binomio\ al\ cuadrado: \\ \\\boxed{(x+y)^{2}=x^{2}+2xy+y^{2} }\\ \ \ \\ \\ ===========================\\ \\ \\ \left(3 \sqrt{x}+ \dfrac{x}{2}\right)^{2} = \\ \\ \\ (3 \sqrt{x})^{2}+\not{2}(3 \sqrt{x} )\left( \dfrac{x}{\not{2}} \right) + \left( \dfrac{x}{2} \right)^{2} = \\ \\ \\ \boxed{9x + 3x\sqrt{x} + \dfrac{ x^{2}}{4}}](https://tex.z-dn.net/?f=Aplicamos%5C+binomio%5C+al%5C+cuadrado%3A+%5C%5C+%5C%5C%5Cboxed%7B%28x%2By%29%5E%7B2%7D%3Dx%5E%7B2%7D%2B2xy%2By%5E%7B2%7D+%7D%5C%5C+%5C+%5C+%5C%5C+%5C%5C+%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%5C%5C+%5C%5C+%5C%5C+%5Cleft%283+%5Csqrt%7Bx%7D%2B+%5Cdfrac%7Bx%7D%7B2%7D%5Cright%29%5E%7B2%7D+%3D+%5C%5C+%5C%5C+%5C%5C+%283+%5Csqrt%7Bx%7D%29%5E%7B2%7D%2B%5Cnot%7B2%7D%283+%5Csqrt%7Bx%7D+%29%5Cleft%28+%5Cdfrac%7Bx%7D%7B%5Cnot%7B2%7D%7D+%5Cright%29+%2B+%5Cleft%28+%5Cdfrac%7Bx%7D%7B2%7D+%5Cright%29%5E%7B2%7D+%3D+%5C%5C+%5C%5C+%5C%5C+%5Cboxed%7B9x+%2B+3x%5Csqrt%7Bx%7D+%2B+%5Cdfrac%7B+x%5E%7B2%7D%7D%7B4%7D%7D)
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