• Matéria: Matemática
  • Autor: NAYARASOUZAJF
  • Perguntado 9 anos atrás

Efetuar as operações:

a) ( 2√5 - 4√7) . (√5 + 2 √7)

b) (4 - √5)²

c) ( 1 - √2)⁴

d) (√20 - √45 + 3√125) : 2√5

Respostas

respondido por: oliverprof
1
a)(2 \sqrt{5} -4 \sqrt{7}) ( \sqrt{5} +2 \sqrt{7})=2.5 +4 \sqrt{35}-4\sqrt{35} -8.7=10-56=-4 \\  \\ b)(4- \sqrt{5} )^2=4^2-2.4. \sqrt{5} +( \sqrt{5} )^2=16-8 \sqrt{5} +5=21-8 \sqrt{5}  \\  \\ c)(1- \sqrt{2} )^4=[(1- \sqrt{2} )^2]^2=[(1^2-2.1. \sqrt{2}+( \sqrt{2} )^2 )]^2= \\ {(1-2 \sqrt{2}+2 )}^2=(3-2 \sqrt{2} )^2=3^2-2.3.2 \sqrt{2} +(2 \sqrt{2})^2  \\  9-12 \sqrt{2} +4.2=17-12 \sqrt{2}   a)(2 \sqrt{5} -4 \sqrt{7}) ( \sqrt{5} +2 \sqrt{7})=2.5 +4 \sqrt{35}-4\sqrt{35} -8.7=10-56=-4 \\  \\ b)(4- \sqrt{5} )^2=4^2-2.4. \sqrt{5} +( \sqrt{5} )^2=16-8 \sqrt{5} +5=21-8 \sqrt{5}  \\  \\ c)(1- \sqrt{2} )^4=[(1- \sqrt{2} )^2]^2=[(1^2-2.1. \sqrt{2}+( \sqrt{2} )^2 )]^2= \\ {(1-2 \sqrt{2}+2 )}^2=(3-2 \sqrt{2} )^2=3^2-2.3.2 \sqrt{2} +(2 \sqrt{2})^2  \\  9-12 \sqrt{2} +4.2=17-12 \sqrt{2}

NAYARASOUZAJF: Obrigada
oliverprof: d)(v20):2v5 - (v45):2v5 +(3v125):2v5
oliverprof: (v4):2 - (v9):2 + (3v25):2
oliverprof: 2:2 -3:2 + (3.5):2
oliverprof: 1- 3:2 +15:2
oliverprof: (2-3+15):2
oliverprof: 14:2=7
oliverprof: Ou vc pode antes resolver o parênteses
oliverprof: Fica (2v5 - 3v5 + 15v5):2v5
oliverprof: 14v5:2v5=7
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