\(1.\left\{{}\begin{matrix}-x+3y=-10\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2y=6\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6}{-2}=-3\\x-5\cdot\left(-3\right)=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+15=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=16-15=1\end{matrix}\right.\\ 2.\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\-2x+8y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\9x=27\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\x=\dfrac{27}{9}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+y=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=7-6=1\\x=3\end{matrix}\right.\) 

\(3.\left\{{}\begin{matrix}3x-5y=-18\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5y=-18\\3x+6y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+2\cdot\left(-3\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=5+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=11\end{matrix}\right.\\ 4.\left\{{}\begin{matrix}4x+3y=-6\\2x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=-6\\4x-10y=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=-38\\2x-5y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x-5\cdot\dfrac{-38}{13}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x+\dfrac{190}{13}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x=16-\dfrac{190}{13}=\dfrac{18}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\x=\dfrac{18}{13}:2=\dfrac{9}{13}\end{matrix}\right.\)

Bài 3:

\(A=75\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\)

Đặt: \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\\ 4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\\ 3B=4^{2005}-1\\ B=\dfrac{4^{2005}-1}{3}\)

\(=>A=75\cdot\dfrac{4^{2005}-1}{3}+25\\ =25\left(4^{2005}-1\right)+25\\ =25\cdot\left(4^{2005}-1+1\right)\\ =25\cdot4^{2005}\\ =25\cdot4\cdot4^{2004}\\ =100\cdot4^{2004}\) 

=> A chia hết cho 100 

Bài 1:

a: \(\dfrac{5\cdot3^{11}+4\cdot3^{17}}{3^9\cdot5^2-3^9\cdot2^3}=\dfrac{3^{11}\cdot\left(5+4\cdot3^6\right)}{3^9\left(5^2-2^3\right)}\)

\(=3^2\cdot\dfrac{5+4\cdot729}{25-8}=3^2\cdot\dfrac{2921}{17}=\dfrac{26289}{17}\)

b: \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{47\cdot50}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{47\cdot50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{1}{3}\cdot\dfrac{24}{50}=\dfrac{24}{150}=\dfrac{8}{50}=\dfrac{4}{25}\)

c: \(1+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=\dfrac{2}{2}+\dfrac{2}{6}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\right)\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)

d: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{99^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-98}{99}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

a: \(\dfrac{-11}{6}< -1\)

\(-1=\dfrac{-9}{9}< \dfrac{8}{-9}\)

Do đó: \(\dfrac{-11}{6}< \dfrac{8}{-9}\)

b: \(-\dfrac{25}{20}< 0\)

\(0< \dfrac{20}{25}\)

Do đó: \(-\dfrac{25}{20}< \dfrac{20}{25}\)

a: \(\dfrac{15}{21}=\dfrac{15:3}{21:3}=\dfrac{5}{7};\dfrac{21}{49}=\dfrac{21:7}{49:7}=\dfrac{3}{7}\)

mà 5>3

nên \(\dfrac{15}{21}>\dfrac{21}{49}\)

b: \(\dfrac{-19}{49}=\dfrac{-19\cdot47}{49\cdot47}=\dfrac{-893}{2303}\)

\(\dfrac{-23}{47}=\dfrac{-23\cdot49}{49\cdot47}=\dfrac{-1127}{2303}\)

mà -893>-1127

nên \(-\dfrac{19}{49}>-\dfrac{23}{47}\)

a: \(\dfrac{998}{555}=\dfrac{555+443}{555}=1+\dfrac{443}{555}\)

\(\dfrac{999}{556}=\dfrac{556+443}{556}=1+\dfrac{443}{556}\)

mà \(\dfrac{443}{555}>\dfrac{443}{556}\)

nên \(\dfrac{998}{555}>\dfrac{999}{556}\)

b: \(\dfrac{315}{380}=1-\dfrac{65}{380};\dfrac{316}{381}=1-\dfrac{65}{381}\)

Ta có: 380<381

=>\(\dfrac{65}{380}>\dfrac{65}{381}\)

=>\(-\dfrac{65}{380}< -\dfrac{65}{381}\)

=>\(-\dfrac{65}{380}+1< -\dfrac{65}{381}+1\)

=>\(\dfrac{315}{380}< \dfrac{316}{381}\)

=>\(-\dfrac{315}{380}>-\dfrac{316}{381}\)

a: \(\dfrac{7}{8}=\dfrac{7\cdot3}{8\cdot3}=\dfrac{21}{24};\dfrac{11}{12}=\dfrac{11\cdot2}{12\cdot2}=\dfrac{22}{24}\)

mà 21<22

nên \(\dfrac{7}{8}< \dfrac{11}{12}\)

b: \(\dfrac{-5}{8}=\dfrac{-5\cdot5}{8\cdot5}=\dfrac{-25}{40};\dfrac{7}{-10}=\dfrac{-7}{10}=\dfrac{-7\cdot4}{10\cdot4}=\dfrac{-28}{40}\)

mà -25>-40

nên \(-\dfrac{5}{8}>\dfrac{7}{-10}\)

a: \(\dfrac{24}{35}=\dfrac{24\cdot6}{35\cdot6}=\dfrac{144}{210};\dfrac{19}{30}=\dfrac{19\cdot7}{30\cdot7}=\dfrac{133}{210}\)

b: \(\dfrac{-9}{21}=\dfrac{-9:3}{21:3}=\dfrac{-3}{7};\dfrac{27}{-63}=\dfrac{27:\left(-9\right)}{-63:\left(-9\right)}=\dfrac{-3}{7}\)

Do đó: \(\dfrac{-9}{21}=\dfrac{27}{-63}\)

a: \(\dfrac{9}{70}=\dfrac{9\cdot3}{70\cdot3}=\dfrac{27}{210};\dfrac{5}{42}=\dfrac{5\cdot5}{42\cdot5}=\dfrac{25}{210}\)

mà 27>25

nên \(\dfrac{9}{70}>\dfrac{5}{42}\)

b: \(\dfrac{-4}{27}=\dfrac{-4\cdot7}{27\cdot7}=\dfrac{-28}{189};\dfrac{15}{-63}=\dfrac{-15}{63}=\dfrac{-15\cdot3}{63\cdot3}=\dfrac{-45}{189}\)

mà -28>-45

nên \(-\dfrac{4}{27}>-\dfrac{15}{63}\)

\(a)\left(\dfrac{2}{3}\right)^5=\dfrac{2^5}{3^5}=\dfrac{32}{243}\\ \left(\dfrac{-2}{3}\right)^5=\dfrac{\left(-2\right)^5}{3^5}=\dfrac{-32}{243}\\ \left(-1\dfrac{3}{4}\right)^2=\left(-\dfrac{7}{4}\right)^2=\dfrac{\left(-7\right)^2}{4^2}=\dfrac{49}{16}\\ \left(-0,1\right)^4=\left(-\dfrac{1}{10}\right)^4=\dfrac{\left(-1\right)^4}{10^4}=\dfrac{1}{10000}\)

\(b)\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\\ \dfrac{790^4}{79^4}=\left(\dfrac{790}{79}\right)^4=10^4=10000\\ \dfrac{3^2}{15^2}=\left(\dfrac{3}{15}\right)^2=\left(\dfrac{1}{5}\right)^2=\dfrac{1^2}{5^2}=\dfrac{1}{25}\\ \dfrac{\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^{n-1}}=\left(-\dfrac{1}{2}\right)^{n-\left(n-1\right)}=\left(-\dfrac{1}{2}\right)^{n-n+1}=\left(-\dfrac{1}{2}\right)\)

a) Ta có: 

\(\dfrac{13}{15}=\dfrac{15-2}{15}=1-\dfrac{2}{15}\)

\(\dfrac{9}{11}=\dfrac{11-2}{11}=1-\dfrac{2}{11}\)

Vì: \(\dfrac{2}{15}< \dfrac{2}{11}=>1-\dfrac{2}{15}>1-\dfrac{2}{11}=>\dfrac{13}{15}>\dfrac{9}{11}\)

b) Ta có: 

\(\dfrac{-9}{17}< 0\)

\(\dfrac{-20}{-21}=\dfrac{20}{21}>0\)

\(=>-\dfrac{9}{17}< 0< \dfrac{20}{21}\)