Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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SOLUCIONARIO DE ARITMETICA DE BALDOR-2. 14 1 5-3. 3 172√14 1 5 = √71 5 = √71√5√71 ∙ 5√5 ∙ 5 = √355√25 = 1 5 √355√3 1 72 = √ 21772 = √217√72EJERCICIO 230√4 1 90 = √361 90 = √361√90 = 19√9019√90 ∙ √10√10 = 19√10√900 = 1930 √10Hallar la raíz cuadrada de las fracciones siguientes mediantereducción a decimal:-1. 5 8-4. 6 120√217 ∙ 2√72 ∙ 2 = √434√144 = 1 12 √434√6 1 20 = √121 20 = √121√20 = 11√20-5. 15 22711√20 ∙ √5√5 = 11√5√100 = 1110 √5 = 1 1 10 √5√15 2 27 = √ 40727 = √407√27√407 ∙ 3√27 ∙ 3√1 221=√81 = 1 √1 2219-2.720-6. 3 625√3 6 25 = √81 25 = √81√25 = 9 5 = 1 4 5-7. 14 117√14 1 17 = √ 23917 = √239√17√239 ∙ 17√17 ∙ 17√4 063=√289 = 1 √4 06317-8. 3 140√3 1 40 = √121 40 = √121√40 = 11√4011√40 ∙ √10√10 = 11√10√400 = 1120 √10-9. 4 190LEONARDO F. APALA TITO 268
SOLUCIONARIO DE ARITMETICA DE BALDOR-4.7406.11800. 1 7 5 0 0 00. 1 3 7 5LEONARDO F. APALA TITO 269
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Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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