Bonjour , voila ma question. Completez les égualites remarquables suivants : [tex] {49t}^{2} ... + 1 = (... - {1)}^{2} [/tex] [tex] {36t}^{2} - ... = (... - 11)
Mathématiques
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Question
Bonjour , voila ma question.
Completez les égualites remarquables suivants :
[tex] {49t}^{2} ... + 1 = (... - {1)}^{2} [/tex]
[tex] {36t}^{2} - ... = (... - 11)(... + ...)[/tex]
[tex]16... = (... + {3t})^{2} [/tex]
[tex] {x}^{2} - ... = (... - \sqrt{3} )(... + ...)[/tex]
[tex] {x}^{2} + ... + ... = (... + { \sqrt{2} })^{2} [/tex]
[tex]... - 4x \sqrt{3 } + ... =( 2x - ... {)}^{2} [/tex]
Completez les égualites remarquables suivants :
[tex] {49t}^{2} ... + 1 = (... - {1)}^{2} [/tex]
[tex] {36t}^{2} - ... = (... - 11)(... + ...)[/tex]
[tex]16... = (... + {3t})^{2} [/tex]
[tex] {x}^{2} - ... = (... - \sqrt{3} )(... + ...)[/tex]
[tex] {x}^{2} + ... + ... = (... + { \sqrt{2} })^{2} [/tex]
[tex]... - 4x \sqrt{3 } + ... =( 2x - ... {)}^{2} [/tex]
1 Réponse
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1. Réponse guestnosdevoir
bonjour
[tex]
{49t}^{2} -14t + 1 = (7t - {1)}^{2}
{36t}^{2} - 121 = (6t - 11)(6t + 6t)
16 t^{2} = (t+ {3t})^{2}
{x}^{2} - 3 = (x - \sqrt{3} )(x + \sqrt{3} )
{x}^{2} + 2 \sqrt{2}x + 2 = (x+ { \sqrt{2} })^{2}
4 x^{2} - 4x \sqrt{3 } + 3=( 2x - \sqrt{ 3} {)}^{2} [/tex]