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\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a) \(2\left(4x-30\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)
\(\Leftrightarrow8x-60-3x+15+4x-40=5x+10\)
\(\Leftrightarrow9x-35=5x+10\)
\(\Leftrightarrow9x-5x=10+35\)
\(\Leftrightarrow4x=45\)
\(\Leftrightarrow x=\dfrac{45}{4}=11,25\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\left(6x+1\right)\)
\(\Leftrightarrow\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}+x=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}-\dfrac{2}{3}=4x-x\)
\(\Leftrightarrow3x=\dfrac{1}{60}\)
\(\Leftrightarrow x=\dfrac{1}{180}\)
c) \(\dfrac{7}{3}-\left(2x-\dfrac{1}{3}\right)=\left(-2\dfrac{1}{6}+1\dfrac{1}{2}\right):0,25\)
\(\Leftrightarrow\dfrac{7}{3}-2x+\dfrac{1}{3}=-1\dfrac{2}{3}:\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-5}{3}.4\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-20}{3}\)
\(\Leftrightarrow2x=\dfrac{8}{3}+\dfrac{20}{3}\)
\(\Leftrightarrow2x=\dfrac{28}{3}\)
\(\Leftrightarrow x=4\dfrac{2}{3}\)
d) \(0,75+\dfrac{5}{9}:x=5\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{5}{9}:x=\dfrac{11}{2}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{11}{2}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{5}{9}:\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{20}{171}\)
3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
\(a)\dfrac{-1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}.\dfrac{152}{11}-\dfrac{1}{4}.\dfrac{68}{11}\)
\(=-38-\dfrac{17}{11}\)
\(=\dfrac{-418}{11}+\dfrac{-17}{11}\)
\(=\dfrac{-435}{11}\)
\(b)\dfrac{31}{9}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{16}{6}+\dfrac{15}{6}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{31}{6}\)
\(\Leftrightarrow\left|x\right|=\dfrac{31}{6}.\dfrac{9}{31}\)
\(\Leftrightarrow\left|x\right|=1,5\)
\(\Leftrightarrow x\in\left\{1,5;-1,5\right\}\)
Vậy \(x\in\left\{1,5;-1,5\right\}\)
\(\dfrac{x-1}{11}\) + \(\dfrac{x-2}{10}\) = \(\dfrac{x-3}{9}\) + \(\dfrac{x-4}{8}\)
\(\dfrac{x-1}{11}\) - 1 + \(\dfrac{x-2}{10}\) - 1 = \(\dfrac{x-3}{9}\) - 1 + \(\dfrac{x-4}{8}\)
\(\dfrac{x-12}{11}\) + \(\dfrac{x-12}{10}\) = \(\dfrac{x-12}{9}\) + \(\dfrac{x-12}{8}\)
(\(x-12\)).( \(\dfrac{1}{11}\) + \(\dfrac{1}{10}\)) = (\(x-12\)) (\(\dfrac{1}{9}\) + \(\dfrac{1}{8}\))
(\(x\) - 12).( \(\dfrac{1}{11}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{9}\) - \(\dfrac{1}{8}\)) = 0
\(x-12\) = 0
\(x\) = 12