\(a,\) Biến đổi \(\left(3x-2\right)\left(3x+1\right)=7\Rightarrow\left[{}\begin{matrix}x=1;y=2\\x=3;y=0\end{matrix}\right.\)

\(b,\) Biến đổi \(5^x.\left(5^{y-x}+1\right)=5^3.26\Rightarrow\left\{{}\begin{matrix}5^x=5^3\\5^{y-x}+1=26\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)

\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)=\dfrac{5}{2}.\dfrac{2}{3}=\dfrac{5}{3}\)

\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)

\(c,C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{210}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.\dfrac{14}{15}.....\dfrac{209}{210}\)

\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.\dfrac{28}{30}....\dfrac{418}{420}=\dfrac{\left(1.4\right)\left(2.5\right)\left(3.6\right)\left(4.7\right).....\left(19.22\right)}{\left(2.3\right)\left(3.4\right)\left(4.5\right)\left(5.6\right)....\left(20.21\right)}\)

\(=\dfrac{\left(1.2.3.......19\right).\left(4.5.6.7.......22\right)}{\left(2.3.4.......20\right).\left(3.4.5.6.......21\right)}=\dfrac{11}{30}\)

\(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+13}=0\Leftrightarrow\left(x-5\right)^{x+1}.\left[1-\left(x-5\right)^{12}\right]=0\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{x+1}=0\\1-\left(x-5\right)^{12}=0\end{matrix}\right.\)

\(\left(x-5\right)^{x+1}=\Leftrightarrow0\left\{{}\begin{matrix}x-5=0\\x+1\ne0\end{matrix}\right.\Leftrightarrow x=5\)

\(1-\left(x-5\right)^{12}=0\Leftrightarrow\left(x-5\right)^{12}=1\Leftrightarrow\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)

Vậy \(x=4;x=5;x=6\)

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2012}.\left(1+2+3+...+2012\right)\)

\(=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+\dfrac{1}{4}.\dfrac{4.5}{2}+...+\dfrac{1}{2012}.\dfrac{2012.2013}{2}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{2013}{2}=\dfrac{1}{2}\left(2+3+4+..2013\right)\)

\(=\dfrac{1}{2}\left(1+2+3+4+...+2013-1\right)=\dfrac{1}{2}\left(\dfrac{2012.2013}{2}-1\right)\)

\(=\dfrac{2025077}{2}\)

Ta có: \(2x+y=z-38\Rightarrow2x+y-z=38\)

Vì \(3x=4y=5z-3x-4y\Rightarrow3x=5z-3x-3x\Rightarrow9x=5z\Rightarrow\dfrac{x}{5}=\dfrac{z}{9}\Rightarrow\dfrac{x}{20}=\dfrac{z}{36}\left(1\right)\)

Vì \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{36}=\dfrac{2x+y-z}{40+15-36}=\dfrac{-38}{19}=-2\Rightarrow\left\{{}\begin{matrix}x=-2.20=-40\\y=-2.15=-30\\z=-2.36=-72\end{matrix}\right.\)

Vậy \(x=-40;y=-30;z=-72\)

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Leftrightarrow\dfrac{abc}{\left(a+b\right)c}=\dfrac{bca}{\left(b+c\right)a}=\dfrac{cab}{\left(c+a\right)b}=\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}\)

\(\Leftrightarrow ac+bc=ab+ac\Leftrightarrow bc=ab\Leftrightarrow a=c\)

Tương tự, chứng minh được \(a=b=c\Rightarrow M=1\)

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)

\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x-1}\right)=\dfrac{49}{99}\)

\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x-1}\right)=\dfrac{49}{99}\Rightarrow1-\dfrac{1}{2x-1}=\dfrac{98}{99}\Rightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)

\(\Rightarrow2x+1=99\Leftrightarrow x=49\)