\(5^{2009}=5^{2000}\cdot5^9\)
Ta có: \(5^{2000}\equiv1\) (\(mod\) \(10000\))
          \(5^9\equiv3125\) (\(mod\) \(10000\))
\(\Rightarrow5^{2000}\cdot5^9\equiv1\cdot3125\) (\(mod\) \(10000\))
\(\Rightarrow5^{2009}\equiv3125\) (\(mod\) \(10000\))
Vậy \(4\) chữ số tận cùng của \(5^{2009}\) là \(3125\) 

Bạn xem lại đề, \(5^2009\) hay \(5^{2009}\)?

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x}=a\\\sqrt[3]{y}=b\end{matrix}\right.\) hệ trở thành \(\left\{{}\begin{matrix}2\left(a^3+b^3\right)=3\left(a^2b+ab^2\right)\\a+b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]=3ab\left(a+b\right)\\a+b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(6^3-18ab\right)=18ab\\a+b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}ab=8\\a+b=6\end{matrix}\right.\)

\(\Rightarrow\) a và b là nghiệm của \(t^2-6t+8=0\Rightarrow\left[{}\begin{matrix}t=4\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=4;b=2\\a=2;b=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=64;y=8\\x=8;y=64\end{matrix}\right.\)

\(x^2-10x^2+9=0\)

\(\Leftrightarrow-9x^2=-9\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{1;-1\right\}\).

- Số lớn nhất \(\Rightarrow x=y=9\), khi đó nó có dạng: \(\overline{19293z}\) chia hết cho 7

\(\Rightarrow\overline{93z}-192\) chia hết cho 7

\(\Rightarrow930+z-192=738+z⋮7\)

\(\Rightarrow z+3⋮7\)

Mà z lớn nhất \(\Rightarrow z=4\)

Vậy số lớn nhất là \(192934\)

- Số nhỏ nhất \(\Rightarrow x=y=0\), khi đó có dạng \(\overline{10203z}\) chia hết cho 7

\(\Rightarrow102-\overline{3z}⋮7\Rightarrow102-\left(30+z\right)⋮7\)

\(\Rightarrow z-2⋮7\), mà z nhỏ nhất \(\Rightarrow z=2\)

Vậy số nhỏ nhất là \(102032\)

n

\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)

Đặt \(n\left(n+2\right).U_n=V_n\Rightarrow V_{n-1}=\left(n-1\right)\left(n+2-1\right).U_{n-1}=\left(n-1\right).\left(n+1\right)U_{n-1}\)

\(\Rightarrow V_n=V_{n-1}\)

\(\Rightarrow V_n=V_{n-1}=V_{n-2}=...=V_1\)

Có \(V_1=1.\left(1+2\right).U_1=1\)

\(\Rightarrow V_n=1\)

\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)

\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)

\(=...\)

\(f\left(x\right)=6x^3-7x^2-16x+m\)

Do \(f\left(x\right)\) chia hết \(2x-5\), theo định lý Bezout:

\(f\left(\dfrac{5}{2}\right)=0\Rightarrow6.\left(\dfrac{5}{2}\right)^3-7.\left(\dfrac{5}{2}\right)^2-16.\left(\dfrac{5}{2}\right)+m=0\)

\(\Rightarrow m=-10\)

Khi đó  \(f\left(x\right)=6x^3-7x^2-16x-10\)

Số dư phép chia cho \(3x-2\):

\(f\left(\dfrac{2}{3}\right)=6.\left(\dfrac{2}{3}\right)^3-7.\left(\dfrac{2}{3}\right)^2-16.\left(\dfrac{2}{3}\right)-10=-22\)

f(x)=6x3−7x2−16x+m

Do $f\left(x\right)$ chia hết $2x-5$, theo định lý Bezout:

�(52)=0⇒6.(52)3−7.(52)2−16.(52)+�=0f(25​)=0⇒6.(25​)3−7.(25​)2−16.(25​)+m=0

$\Rightarrow m=-10$

Khi đó  �(�)=6�3−7�2−16�−10f(x)=6x3−7x2−16x−10

Số dư phép chia cho $3x-2$:

�(23)=6.(23)3−7.(23)2−16.(23)−10=−22f(32​)=6.(32​)3−7.(32​)2−16.(32​)−10=−22

Lời giải:

\(\frac{1719}{3976}=\frac{1}{2+\frac{538}{1719}}=\frac{1}{2+\frac{1}{3+\frac{105}{538}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{13}{105}}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{8+\frac{1}{13}}}}}\)

$\Rightarrow a=8; b=13$

\(\dfrac{1719}{3976}=\dfrac{1}{\dfrac{3976}{1719}}=\dfrac{1}{2+\dfrac{538}{1719}}=\dfrac{1}{2+\dfrac{1}{\dfrac{1719}{538}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{105}{538}}}\)

\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{\dfrac{538}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{13}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{\dfrac{105}{13}}}}}\)

\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{8+\dfrac{1}{13}}}}}\)