Question

a)Tìm số tự nhiên a nhỏ nhất sao cho a chia cho 3, cho 5, cho 7 được số dư thứ tự là 2, 4, 6

b) \(\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)

c)\(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

Answer 1

54444

Answer 2

\(b.\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)

\(\Leftrightarrow\frac{35+9}{105}< \frac{x}{210}< \frac{60+63+35}{105}\)

\(\Leftrightarrow\frac{44}{105}< \frac{x}{210}< \frac{158}{105}\)

\(\Leftrightarrow\frac{88}{210}< \frac{x}{210}< \frac{316}{210}\)

Suy ra \(x\in\left\{89;90;100;...;313;314;315\right\}\)

\(c.\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\frac{1}{11}-\frac{1}{21}-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\frac{21-11-231x+221}{231}=\frac{308}{231}\)

\(\Leftrightarrow-231x=308-21+11-221\)

\(\Leftrightarrow-231x=77\)

\(\Leftrightarrow x=-\frac{77}{231}=-\frac{1}{3}\)

^^