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6 tháng 8 2022
a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)
\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)
b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)
8 tháng 12 2022
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
2 tháng 7 2018
1,
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)
5 tháng 12 2022
\(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=-\dfrac{25}{27}+\dfrac{1}{5}-1=-\dfrac{233}{135}\)
\(D=\dfrac{9}{4}-\dfrac{5}{6}-\dfrac{1}{4}=2-\dfrac{5}{6}=\dfrac{7}{6}\)
G
2 tháng 8 2017
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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NA
2 tháng 8 2017
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17 tháng 12 2017
\(=9.\left(\dfrac{-1}{27}\right)-3.\dfrac{1}{9}+2.\left(\dfrac{-1}{3}\right)+1\)
\(=\left(\dfrac{-1}{3}\right)-\dfrac{1}{3}+\dfrac{-2}{3}+1\)
\(=\dfrac{\left(-1\right)-1+\left(-2\right)+3}{3}\)
\(=\dfrac{-1}{3}\)
7 tháng 12 2017
Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi
1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)
Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)
Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)
\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)
\(P=\dfrac{2+3+4+5+...+17}{2}\)
\(P=\dfrac{152}{2}=76\)
2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)
17 tháng 6 2018
\(25\cdot\left(\dfrac{-1}{5}\right)^3+\dfrac{1}{5}-2\cdot\left(\dfrac{-1}{2}\right)^2-\dfrac{1}{2}\)
\(=25\cdot\left(\dfrac{-1}{125}\right)+\dfrac{1}{5}-2\cdot\left(\dfrac{-1}{4}\right)-\dfrac{1}{2}\)
\(=\left(\dfrac{-1}{5}\right)+\dfrac{1}{5}-\left(\dfrac{-1}{2}\right)-\dfrac{1}{2}\)
\(=0\)
LD
13 tháng 1 2018
a,
\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)
b,
\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
c,
\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)
\(\Rightarrow5\left|x+1\right|^2=180\)
\(\Rightarrow\left|x+1\right|^2=36\)
Mà \(\left|x+1\right|\ge0\)
=> x + 1 = 6 <=> x = 7
\(=\dfrac{1}{3^2}.\dfrac{1}{3}.3^2=\dfrac{1}{3}\)
\(\left(\dfrac{1}{3}\right)^2.\dfrac{1}{3}.9\\ =\dfrac{1}{9}.\dfrac{1}{3}.9\\ =\dfrac{1}{9}.\dfrac{1}{3}.\dfrac{9}{1}\\ =\dfrac{1}{3}.\)