Question

matematica clasa a4a, va rog ajutati-ma

Answer 1

a )

$\boxed{\dfrac{8}{10 } - ( \dfrac{5}{10} + x) = \dfrac{1}{10} } \\ \\ \boxed{ \dfrac{5}{10} + x = \dfrac{8}{10} - \dfrac{1}{10}} \\ \\ \boxed{\dfrac{5}{10} + x = \dfrac{7}{10}} \\ \\ \it\boxed{x = \dfrac{7}{10} - \dfrac{5}{10} = \dfrac{2}{10} }$

b )

$\boxed{2 - \dfrac{2}{2} - y = \dfrac{1}{2} } \\ \\ \boxed{\dfrac{2}{1} ^{ \cdot2} - \dfrac{2}{2} - y = \dfrac{1}{2} } \\ \\ \boxed{ \dfrac{4}{2} - \dfrac{2}{2} - y = \dfrac{1}{2} } \\ \\ \boxed{ \dfrac{2}{2} - y = \dfrac{1}{2} } \\ \\ \it \boxed{ y = \dfrac{2}{2} - \dfrac{1}{2} = \dfrac{1}{2} }$

2 )

$\dfrac{3 + x}{9} < 1 \\ \\ 3 + x < 9 \\ \\ x < 9 - 3 = 6 \\ \\ x \:\epsilon 0,1,2,3,4,5$

$\dfrac{y - 1}{4} < 1 \\ \\ y - 1 < 4 \\ \\ y < 1 + 4 = 5 \\ \\ y = 0...1...2...3...4$

$1 > \dfrac{1}{8} + \dfrac{z}{8} + \dfrac{2}{8} \\ \\ 1 > \dfrac{3}{8} + \dfrac{z}{8} \\ \\ z < 5 \\ \\ z = 0...1...2...3...4$

3 )

$\dfrac{8}{10} - ( \dfrac{9}{10} - x) = \dfrac{4}{10} \\ \\ \dfrac{9}{10} - x = \dfrac{8}{10} - \dfrac{4}{10} \\ \\ \dfrac{9}{10} - x = \dfrac{4}{10} \\ \\ x = \dfrac{9}{10} - \dfrac{4}{10} = \dfrac{5}{10}^{ \div 5} = \dfrac{1}{2}$

4 )

$16 = \dfrac{1}{3} \times x \\ \\ 16 = \dfrac{x}{3} \\ \\ x = 16 \times 3 = 48 \\ \\ 16 = \dfrac{1}{5} \times y \\ \\ 16 = \dfrac{y}{5} \\ \\ y = 16 \times 5 = 80$

5 )

$( \dfrac{9}{2} + \dfrac{5}{2} - 1) - ( \dfrac{1}{4} + \dfrac{3}{4} + 5) = \\ \\ (\dfrac{14}{2} - 1) - ( \dfrac{4}{4} + 5) = \\ \\ (7 - 1) - (1 + 5) = \\ \\ 6 - 6 = 0$

2 )

$x - \dfrac{1}{5} = \dfrac{3}{5} \\ \\ x = \dfrac{1}{5} + \dfrac{3}{5} = \dfrac{4}{5} = > c$

1 )

$(2 + \dfrac{2}{5} + \dfrac{3}{5} ) - ( \dfrac{9}{5} + \dfrac{1}{5} ) + \dfrac{24}{5} = \\ \\ ( \dfrac{10}{5} + \dfrac{2}{5} + \dfrac{3}{5} ) - \dfrac{10}{5} + \dfrac{24}{5} = \\ \\ \dfrac{15}{5} - \dfrac{10}{5} + \dfrac{24}{5} = \dfrac{29}{5}$

Answer 2

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