{"id":10915,"date":"2026-03-22T13:42:00","date_gmt":"2026-03-22T13:42:00","guid":{"rendered":"https:\/\/lereveil224.info\/?post_type=exercices&p=10915"},"modified":"2026-03-22T13:46:19","modified_gmt":"2026-03-22T13:46:19","slug":"exercices-et-corriges-equations-et-inequations","status":"publish","type":"exercices","link":"https:\/\/lereveil224.info\/index.php\/exercices\/exercices-et-corriges-equations-et-inequations\/","title":{"rendered":"Exercices et Corrig\u00e9s : \u00c9quations et In\u00e9quations"},"content":{"rendered":"\n

Partie I : \u00c9quations<\/h1>\n\n\n\n

1. Montrer que dans chacun des cas suivants les \u00e9quations (E1) et (E2) dans A sont \u00e9quivalentes.

1) (E1) : x\/2 – 3 = x\/3 – 4
   (E2) : 3x – 18 = 2x – 24 ; A = \u211d

2) (E1) : |x – 2| = 5
   (E2) : (x – 2)\u00b2 = 25 ; A = \u211d

3) (E1) : x + x\u00b3 = 4x
   (E2) : 1 + x\u00b2 = 4 ; A = \u211d

4) (E1) : (2x + 1)\/(x – 1) = (x – 1)\/(2x + 1)
   (E2) : (2x + 1)\u00b2 = (x – 1)\u00b2 ; A = \u211d \\ {1 ; -1\/2}

<\/p>\n\n\n\n

Partie II : In\u00e9quations<\/h1>\n\n\n\n

2. Dans chacun des cas suivants, les in\u00e9quations (I1) et (I2) dans A sont-elles \u00e9quivalentes ?

1) (I1) : x\/2 – 3 \u2264 x\/3 – 4
   (I2) : 3x – 18 \u2264 2x – 24 ; A = \u211d

2) (I1) : |x – 2| \u2264 5
   (I2) : (x – 2)\u00b2 \u2264 25 ; A = \u211d

3) (I1) : x + x\u00b3 \u2264 4x
   (I2) : 1 + x\u00b2 \u2264 4 ; A = \u211d

4) (I1) : (2x + 1)\/(x – 1) \u2264 (x – 1)\/(2x + 1)
   (I2) : (2x + 1)\u00b2 \u2264 (x – 1)\u00b2 ; A = \u211d \\ {1 ; -1\/2}

<\/p>\n\n\n\n

Corrig\u00e9s<\/h1>\n\n\n\n

Equations :<\/h2>\n\n\n\n

1) (E1) \u21d4 x\/2 – 3 = x\/3 – 4
\u00a0\u00a0 Multiplier par 6 :
\u00a0\u00a0 3x – 18 = 2x – 24 \u21d4 (E2)
\u00a0\u00a0 Donc \u00e9quivalentes.

2) |x – 2| = 5 \u21d4 x – 2 = 5 ou x – 2 = -5
\u00a0\u00a0 \u21d2 x = 7 ou x = -3
\u00a0\u00a0 (x – 2)\u00b2 = 25 \u21d4 m\u00eames solutions
\u00a0\u00a0 Donc \u00e9quivalentes.

3) x + x\u00b3 = 4x \u21d4 x\u00b3 – 3x = 0
\u00a0\u00a0 \u21d4 x(x\u00b2 – 3) = 0
\u00a0\u00a0 \u21d2 x = 0 ou x = \u00b1\u221a3

\u00a0\u00a0 1 + x\u00b2 = 4 \u21d4 x\u00b2 = 3
\u00a0\u00a0 \u21d2 x = \u00b1\u221a3

\u00a0\u00a0 Pas totalement \u00e9quivalentes (solution 0 en plus dans E1).

4) (2x + 1)\/(x – 1) = (x – 1)\/(2x + 1)
\u00a0\u00a0 \u21d4 (2x + 1)\u00b2 = (x – 1)\u00b2 avec conditions x \u2260 1 et x \u2260 -1\/2
\u00a0\u00a0 Donc \u00e9quivalentes sur A.<\/p>\n\n\n\n

In\u00e9quations :<\/strong><\/h2>\n\n\n\n

1) M\u00eame transformation \u21d2 \u00e9quivalentes.

2) |x – 2| \u2264 5 \u21d4 -5 \u2264 x – 2 \u2264 5 \u21d2 -3 \u2264 x \u2264 7
\u00a0\u00a0 (x – 2)\u00b2 \u2264 25 \u21d2 m\u00eame r\u00e9sultat \u21d2 \u00e9quivalentes.

3) x + x\u00b3 \u2264 4x \u21d4 x\u00b3 – 3x \u2264 0 \u21d4 x(x\u00b2 – 3) \u2264 0
\u00a0\u00a0 \u21d2 solutions diff\u00e9rentes de 1 + x\u00b2 \u2264 4
\u00a0\u00a0 Donc non \u00e9quivalentes.

4) M\u00eame transformation avec conditions \u21d2 \u00e9quivalentes.<\/p>\n","protected":false},"featured_media":0,"template":"","classes":[64,63],"class_list":["post-10915","exercices","type-exercices","status-publish","hentry","classe-11e-se","classe-11e-sm"],"acf":[],"_links":{"self":[{"href":"https:\/\/lereveil224.info\/index.php\/wp-json\/wp\/v2\/exercices\/10915","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/lereveil224.info\/index.php\/wp-json\/wp\/v2\/exercices"}],"about":[{"href":"https:\/\/lereveil224.info\/index.php\/wp-json\/wp\/v2\/types\/exercices"}],"wp:attachment":[{"href":"https:\/\/lereveil224.info\/index.php\/wp-json\/wp\/v2\/media?parent=10915"}],"wp:term":[{"taxonomy":"classe","embeddable":true,"href":"https:\/\/lereveil224.info\/index.php\/wp-json\/wp\/v2\/classes?post=10915"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}