Question

4supra19×(1supra3+1supra4)=; (2supra5×3antregi1supra3-1)×1antreg 4supra5+1supra10=;va rog ajutătoare și pe mine​

Answer 1

Răspuns: Ai rezolvarea mai jos

Explicație pas cu pas:

Salutare!

$\bf \bigstar~\dfrac{4}{19} \cdot\bigg(\dfrac{1}{3}+\dfrac{1}{4}\bigg)=\bf \dfrac{4}{19} \cdot\bigg(\!{^{^{^{^{^{^{\displaystyle\bf 4)}}}}}}}\!\!\!\dfrac{1}{3}+\!{^{^{^{^{^{^{\displaystyle\bf 3)}}}}}}}\!\!\!\dfrac{1}{4}\bigg)=\dfrac{4}{19} \cdot\dfrac{4+3}{12}=\dfrac{\not4}{19} \cdot\dfrac{7}{\not12}=\boxed{\boxed{\bf \dfrac{7}{57}}}$

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$\bf\bigstar~ \bigg(\dfrac{2}{5}\cdot3\dfrac{1}{3}-1\bigg)\cdot 1\dfrac{4}{5}+\dfrac{1}{10}= \bigg(\dfrac{2}{5}\cdot\dfrac{3\cdot3+1}{3}-\dfrac{1}{1} \bigg)\cdot \dfrac{1\cdot5+4}{5}+\dfrac{1}{10}=$

$\it~~$

$\bf \bigg(\dfrac{2}{\not5}\cdot\dfrac{\not10}{3}-\dfrac{1}{1} \bigg)\cdot \dfrac{9}{5}+\dfrac{1}{10}= \bigg(\dfrac{2}{1}\cdot\dfrac{2}{3}-\dfrac{1}{1} \bigg)\cdot \dfrac{9}{5}+\dfrac{1}{10}=\bigg(\dfrac{4}{3}-\!{^{^{^{^{^{^{\displaystyle\bf 3)}}}}}}}\!\!\!\dfrac{1}{1} \bigg)\cdot \dfrac{9}{5}+\dfrac{1}{10}=$

$\it~~~$

$\bf \dfrac{4-3}{3}\cdot \dfrac{9}{5}+\dfrac{1}{10}=\dfrac{1}{\not3}\cdot \dfrac{\not9}{5}+\dfrac{1}{10}=\dfrac{1}{1}\cdot \dfrac{3}{5}+\dfrac{1}{10}={^{^{^{^{^{^{\displaystyle\bf 2)}}}}}}}\!\!\!\dfrac{3}{5}+\dfrac{1}{10}=\dfrac{6+1}{10}=\boxed{\boxed{\bf \dfrac{7}{10}}}$

$\it~~$

PS: Te rog sa glisezi spre dreapta pentru a vedea toată rezolvarea

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