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Bài làm :
1) \(\frac{x+11}{4}=\frac{2x+4}{5}\)
\(\Leftrightarrow\left(x+11\right).5=4.\left(2x+4\right)\)
\(\Leftrightarrow5x+55=8x+16\)
\(\Leftrightarrow5x-8x=16-55\)
\(\Leftrightarrow-3x=-39\)
\(\Leftrightarrow x=\frac{-39}{-3}=\frac{39}{3}=13\)
2)\(\frac{x+4}{x+10}=\frac{3}{5}\)
\(\Leftrightarrow\left(x+4\right).5=\left(x+10\right).3\)
\(\Leftrightarrow5x+20=3x+30\)
\(\Leftrightarrow5x-3x=30-20\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=\frac{10}{2}=5\)
3)\(\frac{x+8}{x+14}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+8\right).3=\left(x+14\right).2\)
\(\Leftrightarrow3x+24=2x+28\)
\(\Leftrightarrow3x-2x=28-24\)
\(\Leftrightarrow x=4\)
a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)
\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)
=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)
=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)
b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)
=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)
c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)
\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)
\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)
=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)
\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)
Chúc bạn học tốt 
\(2^{3x+5}+5.2^{x+1}.2^{2x+3}+2^{3x}=904\)
\(\Leftrightarrow2^{3x}.2^5+5.2^{3x}.2^4+2^{3x}=904\)
\(\Leftrightarrow2^{3x}\left(2^5+5.2^4+1\right)=904\)
\(\Leftrightarrow8^x\left(32+80+1\right)=904\)
\(\Leftrightarrow8^x.113=904\)
\(\Leftrightarrow8^x=8\)
\(\Leftrightarrow x=1\)
V...
\(5+5^2+5^3...+5^{96}\) Biến đổi phép tính một chút cho đơn giản ta được:
\(5+5^2+5^3+...+5^{96}\Leftrightarrow1+1^2+1^3+...+1^{96}\)
Ta có: \(1^{96}=1\)mà . Ta lại có:
\(1+1^2+1^3+...+1^{96}=1+1+1+...+1\)
96 chữ số 1 hay tổng trên là 96
Mà \(96⋮96\Rightarrow1+1^2+1^3+...+1^{96}\)hay \(5+5^2+5^3+...+5^{96}⋮96\RightarrowĐPCM\)
1. \(n^2+n-17\)là bội của n+5\(\Leftrightarrow\)\(n^2+n-17\)chia hết n+5
Ta có \(n^2+n-17⋮n+5\)
\(\Rightarrow n^2+n+5-22⋮n+5\)
\(\Rightarrow22⋮n+5\)
\(\Rightarrow n+5\inƯ\left(22\right)\)
\(\Rightarrow n+5\in\left(1;2;11;22;-1;-2;-11;-22\right)\)
\(\Rightarrow n\in\left(-4;-3;6;17;-6;-7;-16;-27\right)\)
Ta thấy 36 là BCNN( 18, 12, 9, 4) nên ta có:
\(\frac{1}{18}=\frac{1\times2}{18\times2}=\frac{2}{36};\frac{x}{12}=\frac{x\times3}{12\times3}=\frac{x\times3}{36};\frac{y}{9}=\frac{y\times4}{9\times4}=\frac{y\times4}{36};\frac{1}{4}=\frac{1\times9}{4\times9}=\frac{9}{36}\)\(=\frac{9}{36}\)
quy đồng xong ta có
\(\frac{2}{36}< \frac{x\times3}{36}< \frac{y\times4}{36}< \frac{9}{36}\)
để thoã mãn điều kiện trên vậy x=1;y=2
\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}.\left(-3\right)^2\)
\(=\dfrac{2}{3}+\dfrac{5}{6}.\dfrac{1}{5}-\dfrac{1}{18}.9\)
\(=\dfrac{2}{3}+\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{4}{6}+\dfrac{1}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}\cdot\left(-3\right)^2\)
\(=\dfrac{4}{6}+\dfrac{5}{6}:5-\dfrac{1}{18}\times9\)
\(=\dfrac{4}{6}+\dfrac{5}{6}\times\dfrac{1}{5}-\dfrac{1}{18}\times\dfrac{9}{1}\)
\(=\dfrac{4}{6}+\dfrac{5}{30}-\dfrac{9}{18}\)
\(=\dfrac{20}{30}+\dfrac{5}{30}-\dfrac{1}{2}\)
\(=\dfrac{25}{30}-\dfrac{1}{2}\)
\(=\dfrac{25}{30}-\dfrac{15}{30}\)
\(\dfrac{10}{30}=\dfrac{1}{3}\)
\(5^{3x-2}+3\dfrac{1}{2}\cdot5^5=28,5\cdot5^5\)
=>\(5^{3x-2}=28,5\cdot5^5-3,5\cdot5^5\)
=>\(5^{3x-2}=5^5\cdot25=5^7\)
=>3x-2=7
=>3x=9
=>x=3