Задачи с дроб

[Гост]

IMG_20241201_150229.jpg (47.65 KiB) Прегледано 372 пъти

Задача 10 и 11 .Благодаря.

[ammornil]

$\boxed{10}\\[12pt] \left(3\frac{1}{2}+\frac{1}{3}\right)-\frac{11}{2}=\left(\underbrace{\frac{\overset{3}{7}}{2}+\frac{\overset{2}{1}}{3}}_{6}\right)-\frac{11}{2}=\frac{7\cdot{}3+2\cdot{}1}{6}-\frac{11}{2}=\underbrace{\frac{\overset{1}{23}}{6}-\frac{\overset{3}{11}}{2}}_{6}=\frac{23-33}{6}=-\frac{10}{6}=-\frac{5}{3} \\[6pt] 5\frac{3}{8}+\left(-\frac{5}{3}\right)=\underbrace{\frac{\overset{3}{43}}{8}-\frac{\overset{8}{5}}{3}}_{24}=\frac{3\cdot{43}-8\cdot{5}}{24}=\frac{89}{24}=3\frac{17}{24}$ $$\text{Отг.:}\quad 3\frac{17}{24}$$

[ammornil]

$ \boxed{11}\\[12pt] \text{ден }1: \quad 4\frac{18}{25}[t]=\frac{118}{25}[t] \\[6pt] \text{ден }2: \quad \underbrace{\frac{\overset{1}{118}}{25}-\frac{\overset{5}{8}}{5}}_{25}=\frac{1\cdot{118}-5\cdot{8}}{25}=\frac{78}{25}=3\frac{3}{25}[t] \\[6pt] \hspace{3em} \text{(а)} \quad \frac{118}{25}+\frac{78}{25} = \frac{196}{25}= 7\frac{21}{25}[t] \\[6pt] \text{ден }3:\quad \underbrace{\frac{\overset{25}{13}}{1}-\frac{\overset{1}{196}}{25}}_{25}=\frac{25\cdot{13}-1\cdot{196}}{25}=\frac{129}{25}=5\frac{4}{25}[t] \\[6pt] \hspace{3em} \text{(б)} \quad \frac{129}{25}-\frac{78}{25} = \frac{51}{25}= 2\frac{1}{25}[t] $ $$ \text{Отг.}: \quad \text{(а)} 7\frac{21}{25}[t], \quad \text{(б)} 2\frac{1}{25}[t] $$

Проверете сметките за изчислителни грешки, защото смятах "на ум" и директно в LaTeX.

[Гост]

$\boxed{10}\\[12pt] \left(3\frac{1}{2}+\frac{1}{3}\right)-\frac{11}{2}=\left(\underbrace{\frac{\overset{3}{7}}{2}+\frac{\overset{2}{1}}{3}}_{6}\right)-\frac{11}{2}=\frac{7\cdot{}3+2\cdot{}1}{6}-\frac{11}{2}=\underbrace{\frac{\overset{1}{23}}{6}-\frac{\overset{3}{11}}{2}}_{6}=\frac{23-33}{6}=-\frac{10}{6}=-\frac{5}{3} \\[6pt] 5\frac{3}{8}+\left(-\frac{5}{3}\right)=\underbrace{\frac{\overset{3}{43}}{8}-\frac{\overset{8}{5}}{3}}_{24}=\frac{3\cdot{43}-8\cdot{5}}{24}=\frac{89}{24}=3\frac{17}{24}$ $$\text{Отг.:}\quad 3\frac{17}{24}$$

Има грешка в задачата .Дадено е ,че дробта е 11/12 а ,не 11/2

[ammornil]

Благодаря за корекцията. Можеше да дадете и вярното решение...
$\boxed{10}\\[12pt] \left(3\frac{1}{2}+\frac{1}{3}\right)-\frac{11}{12}=\left(\underbrace{\frac{\overset{3}{7}}{2}+\frac{\overset{2}{1}}{3}}_{6}\right)-\frac{11}{12}=\frac{7\cdot{}3+2\cdot{}1}{6}-\frac{11}{12}=\underbrace{\frac{\overset{2}{23}}{6}-\frac{\overset{1}{11}}{12}}_{12}=\frac{2\cdot{}23-11}{12}=\frac{35}{12}=2\frac{11}{12} \\[6pt] 5\frac{3}{8}+\frac{35}{12}=\underbrace{\frac{\overset{3}{43}}{8}+\frac{\overset{2}{35}}{12}}_{24}=\frac{3\cdot{43}+2\cdot{35}}{24}=\frac{129+70}{24}=\frac{199}{24}=8\frac{7}{24}$ $$\text{Отг.:}\quad 8\frac{7}{24}$$