Explicación paso a paso:
∆ Solución 01:
[tex]9 + ( \: ) + 49 {x}^{2} = {(a + b)}^{2} \\ 9 + ( \: ) + 49 {x}^{2} = {a}^{2} + (2ab) + {b}^{2} \\ \\ igualamos: \\ 9 = {a}^{2} \\ 3 = a \\ a = 3 \\ \\ 49 {x}^{2} = {b}^{2} \\ 7x = b \\ b = 7x \\ \\ lo \: que \: va \: dentro \: del \: paréntesis \: es \: 2ab: \\ \\ calculamos \: 2ab \\ \\ 2ab = 2(3)(7x) \\ 2ab = 42x \\ \\ respuesta: 42x[/tex]
∆ Solución 02:
[tex] {x}^{4} + 20 {x}^{2} y + ( \: ) = {(a + b)}^{2} \\ {x}^{4} + 20 {x}^{2} y + ( \: ) = {a}^{2} + 2ab + {b}^{2} \\ \\ {x}^{4} = {a}^{2} \\ {x}^{2} = a \\ a = {x}^{2} \\ \\ 20 {x}^{2} y = 2ab \\ 10 {x}^{2} y = ab \\ 10y = b \\ \\ nos \: pide \: calcular \: {b}^{2} : \\ {b}^{2} = {(10y)}^{2} \\ {b}^{2} = 100 {y}^{2} \\ \\ el \: valor \: que \: falta \: es: \: 100 {y}^{2} [/tex]
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Explicación paso a paso:
∆ Solución 01:
[tex]9 + ( \: ) + 49 {x}^{2} = {(a + b)}^{2} \\ 9 + ( \: ) + 49 {x}^{2} = {a}^{2} + (2ab) + {b}^{2} \\ \\ igualamos: \\ 9 = {a}^{2} \\ 3 = a \\ a = 3 \\ \\ 49 {x}^{2} = {b}^{2} \\ 7x = b \\ b = 7x \\ \\ lo \: que \: va \: dentro \: del \: paréntesis \: es \: 2ab: \\ \\ calculamos \: 2ab \\ \\ 2ab = 2(3)(7x) \\ 2ab = 42x \\ \\ respuesta: 42x[/tex]
∆ Solución 02:
[tex] {x}^{4} + 20 {x}^{2} y + ( \: ) = {(a + b)}^{2} \\ {x}^{4} + 20 {x}^{2} y + ( \: ) = {a}^{2} + 2ab + {b}^{2} \\ \\ {x}^{4} = {a}^{2} \\ {x}^{2} = a \\ a = {x}^{2} \\ \\ 20 {x}^{2} y = 2ab \\ 10 {x}^{2} y = ab \\ 10y = b \\ \\ nos \: pide \: calcular \: {b}^{2} : \\ {b}^{2} = {(10y)}^{2} \\ {b}^{2} = 100 {y}^{2} \\ \\ el \: valor \: que \: falta \: es: \: 100 {y}^{2} [/tex]