\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=12\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\Rightarrow x=\frac{3}{11}+\frac{8}{11}=\frac{11}{11}=1\)

\(x-\left(\frac{20}{11\cdot13}+\frac{20}{13\cdot15}+...+\frac{20}{53\cdot55}\right)=\frac{3}{11}\)

Đặt \(x-A=\frac{3}{11}\)

\(\frac{A}{10}=\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\)

\(\frac{A}{10}=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)

\(\frac{A}{10}=\frac{1}{11}-\frac{1}{55}\)

\(\frac{A}{10}=\frac{4}{55}\)

\(A=\frac{8}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}\)

\(\Rightarrow x=1\)

đầu bài sai rùi !!!

(chắc chắn một tỉ phần trăm)

\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{15}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}\cdot\left(3^5+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=\frac{2\cdot248}{3\cdot5}\)

\(=\frac{496}{15}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(1\right)\)

\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

Từ (1) và (2) => đpcm

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