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leopard2
@leopard2
January 2021
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aidez-moi svp c'est pour demain !!!
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Pour tout x≥0 démontrer que [ √x-1] ≤ √[x-1] [ ] signifie valeur absolue
(√x-1)(√x+1)= x-1 ⇒ √x - 1 = (x-1)/(√x + 1 )
x≥ 0 ⇒ √x+1 ≥ 0 √x-1 ≤ √x+1 ⇒ (√x-1) / (√x+1) ≤1 ⇒ (x-1)(√x-1)/(√x+1)≤(x-1)
(√x - 1 )² = ( √x - 1) ( √x-1) = (√x-1)(x-1)/(√x+1)
⇒ ( √x-1)² = (x-1) * (√x-1)/(√x+1)
⇒ ( √x-1)² ≤ ( x-1) ⇒ [(√x-1)²] ≤ [x-1]
⇒ [√x-1]²≤ [x-1]
⇒ [√x-1] ≤ √[x-1]
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leopard2
January 2021 | 0 Respostas
aidez-moi svp c'est pour demain !!!
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Leopard2
April 2019 | 0 Respostas
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Leopard2
April 2019 | 0 Respostas
Merci de votre aide je bute sur cet exercice ! c'est pour lundi ! :)))
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(√x-1)(√x+1)= x-1 ⇒ √x - 1 = (x-1)/(√x + 1 )
x≥ 0 ⇒ √x+1 ≥ 0 √x-1 ≤ √x+1 ⇒ (√x-1) / (√x+1) ≤1 ⇒ (x-1)(√x-1)/(√x+1)≤(x-1)
(√x - 1 )² = ( √x - 1) ( √x-1) = (√x-1)(x-1)/(√x+1)
⇒ ( √x-1)² = (x-1) * (√x-1)/(√x+1)
⇒ ( √x-1)² ≤ ( x-1) ⇒ [(√x-1)²] ≤ [x-1]
⇒ [√x-1]²≤ [x-1]
⇒ [√x-1] ≤ √[x-1]