Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{46p}{p}+\dfrac{46q}{q}=\dfrac{46p}{p}.\dfrac{46q}{q}\)
\(=\dfrac{46p+46q}{p.q}=\dfrac{46.46}{p.q}=46.\left[q+p\right]=46.46\)\(=p+q=46\)
vậy các cặp số nguyên tố thỏa mãn là:
[23,23];[3,43];[5,41];[17,29]
\(=\dfrac{34}{11}\cdot\dfrac{23}{17}\cdot\dfrac{27}{46}\cdot\dfrac{22}{9}=\dfrac{34}{17}\cdot\dfrac{23}{11}\cdot\dfrac{22}{46}\cdot\dfrac{27}{9}\)
\(=2\cdot3\cdot\dfrac{1}{2}\cdot2=6\)
P=\(\dfrac{1}{1\cdot2}\)+\(\dfrac{2}{2\cdot4}\)+\(\dfrac{3}{4\cdot7}\)+...+\(\dfrac{10}{46\cdot56}\)
\(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{56}\)
\(=1-\dfrac{1}{56}=\dfrac{55}{56}\)
a,\(\dfrac{3}{7}\).\(\dfrac{14}{5}\)=\(\dfrac{6}{5}\)
b,\(\dfrac{35}{9}\).\(\dfrac{81}{7}\)=45
c,\(\dfrac{28}{17}\).\(\dfrac{68}{14}\)=8
d,\(\dfrac{35}{46}\).\(\dfrac{23}{105}\)=\(\dfrac{1}{6}\)
e,\(\dfrac{12}{5}\):\(\dfrac{16}{15}\)=\(\dfrac{12}{5}\).\(\dfrac{15}{16}\)=\(\dfrac{9}{4}\)
i,\(\dfrac{9}{8}\):\(\dfrac{6}{5}\)=\(\dfrac{9}{8}\).\(\dfrac{5}{6}\)=\(\dfrac{15}{16}\)
Lời giải:
Ta có:
\(\frac{46}{P}+\frac{46}{Q}=\frac{46}{P}.\frac{46}{Q}\)
\(\Leftrightarrow \frac{1}{P}+\frac{1}{Q}=\frac{46}{P.Q}\)
\(\Leftrightarrow \frac{P+Q}{P.Q}=\frac{46}{P.Q}\Leftrightarrow P+Q=46\)
Không mất tổng quát giả sử \(P\geq Q\Rightarrow 46=P+Q\geq 2Q\)
\(\Leftrightarrow Q\leq 23\).
Vì \(Q\in\mathbb{P}\Rightarrow Q\in\left\{2;3;5;7;11;13;17;19;23\right\}\)
\(\Rightarrow P\in\left\{44;43;41;39;35;33;29;27;23\right\}\) (theo thứ tự)
Mà \(P\in\mathbb{P}\Rightarrow P\in\left\{43;41;29;23\right\}\)
Vậy các bộ số (P,Q) thỏa mãn là:
\(\left\{(3;43);(5;41);(17;29);(23,23)\right\}\) và hoán vị từng bộ.