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.

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

\(\left\{{}\begin{matrix}99^{29}=\left(99^2\right)^{10}=9801^{10}\\9999^{10}=9999^{10}\end{matrix}\right.\)

\(9801^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)

\(\left\{{}\begin{matrix}\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\\\left(-\dfrac{25}{46}\right)^{2005}< 0\end{matrix}\right.\)

\(\Rightarrow\left(-\dfrac{25}{46}\right)^{2005}< \left|-\dfrac{25}{46}\right|\)

a/ Ta có :

\(99^{20}=\left(99^2\right)^{10}=9081^{10}\)

Vì \(9081^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)

b/ Ta có :

\(\left|\dfrac{-25}{46}\right|=\dfrac{25}{46}>0\)

\(\left(\dfrac{-25}{46}\right)^{2005}< 0\)

\(\Leftrightarrow\left|\dfrac{-25}{46}\right|>\left(\dfrac{-25}{46}\right)^{2005}\)

a) Ta có: \(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)

\(9999^{10}=\left(99.101\right)^{10}\)

Rõ ràng ta thấy:

\(99.99< 99.101\)

Nên \(\left(99.99\right)^{10}< \left(99.101\right)^{10}\)

Vậy \(99^{20}< 9999^{10}\)

b) Ta có: \(\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\)

\(\left(-\dfrac{25}{46}\right)^{2005}< 0\) (theo tính chất số mũ của số hữu tỉ)

Vậy \(\left|-\dfrac{25}{46}\right|>\left(-\dfrac{25}{46}\right)^{2005}\)

\(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{56}\)

\(=1-\dfrac{1}{56}=\dfrac{55}{56}\)

\(=\dfrac{34}{11}\cdot\dfrac{23}{17}\cdot\dfrac{27}{46}\cdot\dfrac{22}{9}=\dfrac{34}{17}\cdot\dfrac{23}{11}\cdot\dfrac{22}{46}\cdot\dfrac{27}{9}\)

\(=2\cdot3\cdot\dfrac{1}{2}\cdot2=6\)

tìm x,y z ak bn

\(\dfrac{x+46}{20}=x\dfrac{2}{5}\)

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)

\(\Leftrightarrow5\cdot\left(x+46\right)=20\cdot\left(5x+2\right)\)

\(\Leftrightarrow100x+40-5x-230=0\)

\(\Leftrightarrow95x-190=0\)

\(\Leftrightarrow x=2\)