Выражение: 1) 5/8х -(1/4х-1/12у)+1/3у 2)9,4х+(2х-11 - вопрос №5843108158141121329421134359321671 от katy54445 06.01.2023 15:55

Question

Математика: Выражение: 1) 5/8х -(1/4х-1/12у)+1/3у 2)9,4х+(2х-11 3/4у)-3 5/9у 3)2 1/6х-(7х-1 3/4у)+2 1/4у 4)3,5х+(6 1/4х-7у)-7у

katy54445 · Answer

1) ⁵/₈х -(¹/₄х-¹/₁₂у)+¹/₃у=⁵/₈х - ¹/₄х + ¹/₁₂у +¹/₃у= ³/₈x + ⁵/₁₂y
⁵/₈х - ¹/₄х = ⁵/₈х - ¹ˣ²/₄х =⁵/₈х - ²/₈х = ⁵⁻²/₈x = ³/₈x
¹/₁₂у +¹/₃у=¹/₁₂у +¹ˣ⁴/₁₂у=¹/₁₂у +⁴/₁₂у= ¹⁺⁴/₁₂y = ⁵/₁₂y

2)9,4х+(2х-11³/₄у)-3⁵/₉у = 9,4х+2х-11³/₄у-3⁵/₉у = 11,4x +8⁷/₃₆y
9,4х+2х=11,4x
11³/₄у-3⁵/₉у = ⁴⁷/₄y - ³²/₉y = ⁴⁷ˣ⁹/₃₆y - ³²ˣ⁴/₃₆y = ⁴²³/₃₆y - ¹²⁸/₃₆y = ⁴²³⁻¹²⁸/₃₆y=
²⁹⁵/₃₆y = 8⁷/₃₆y

3)2 ¹/₆х-(7х-1³/₄у)+2¹/₄у= 4y - 4⁵/₆x
2¹/₆х-7х = 2¹/₆х - 6⁶/₆х = ¹/₆x - 4⁶/₆x = - 4⁵/₆x
1³/₄у+2¹/₄у=3³⁺¹/₄у = 3⁴/₄y = 4y

4)3,5х+(6¹/₄х-7у)-7у = 3,5х+6¹/₄х-7у-7у = 6x-14y
3,5х+6¹/₄х = ³⁵/₁₀х+²⁵/₄х = ³⁵ˣ²/₂₀х+²⁵ˣ⁵/₂₀х = ⁷⁰/₂₀х+⁵⁰/₂₀х = ⁷⁰⁺⁵⁰/₂₀x=¹²⁰/₂₀x=6x