Tìm tất cả các nguyên tố P và Q thõa mãn
\(\dfrac{46}{P}\) + \(\dfrac{46}{Q}\) = \(\dfrac{46}{P}\) . \(\dfrac{46}{Q}\)
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\(\dfrac{46p}{p}+\dfrac{46q}{q}=\dfrac{46p}{p}.\dfrac{46q}{q}\)
\(=\dfrac{46p+46q}{p.q}=\dfrac{46.46}{p.q}=46.\left[q+p\right]=46.46\)\(=p+q=46\)
vậy các cặp số nguyên tố thỏa mãn là:
[23,23];[3,43];[5,41];[17,29]
1/Ta co :
\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}\right)+\dfrac{46}{45}.\left(\dfrac{15}{7}-\dfrac{45}{46}\right)\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-\dfrac{15}{7}.\dfrac{46}{45}+\dfrac{46}{45}.\dfrac{15}{7}-\dfrac{46}{45}.\dfrac{45}{46}\)
=\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}+\dfrac{46}{45}\right)-1\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-1\)
=\(\dfrac{3}{7}-1=\dfrac{-4}{7}\)
2/Ta co
\(\dfrac{43}{47}.\left(\dfrac{18}{37}+\dfrac{47}{43}\right)-\dfrac{18}{3}.\left(\dfrac{43}{47}+\dfrac{37}{36}\right)\)
=\(\dfrac{43}{47}.\dfrac{18}{37}+\dfrac{43}{47}.\dfrac{47}{43}-\dfrac{18}{37}.\dfrac{43}{47}+\dfrac{18}{37}.\dfrac{37}{36}\)
=\(\dfrac{18}{37}.\left(\dfrac{43}{37}-\dfrac{43}{37}+\dfrac{37}{36}\right)+1\)
=\(\dfrac{18}{37}.\dfrac{37}{36}+1\)
=\(\dfrac{1}{2}+1=\dfrac{3}{2}\)
tick cho mk nha
P=\(\dfrac{1}{1\cdot2}\)+\(\dfrac{2}{2\cdot4}\)+\(\dfrac{3}{4\cdot7}\)+...+\(\dfrac{10}{46\cdot56}\)
\(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{56}\)
\(=1-\dfrac{1}{56}=\dfrac{55}{56}\)
\(=\dfrac{34}{11}\cdot\dfrac{23}{17}\cdot\dfrac{27}{46}\cdot\dfrac{22}{9}=\dfrac{34}{17}\cdot\dfrac{23}{11}\cdot\dfrac{22}{46}\cdot\dfrac{27}{9}\)
\(=2\cdot3\cdot\dfrac{1}{2}\cdot2=6\)
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.