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Bài 2:
a: \(=\dfrac{-2}{3}\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\dfrac{2}{3}=\dfrac{-2}{3}+\dfrac{2}{3}=0\)
b: \(=\dfrac{3}{4}\left(\dfrac{9}{13}+\dfrac{7}{13}-\dfrac{3}{13}\right)=\dfrac{3}{4}\)
c: \(=\dfrac{-7}{12}\left(\dfrac{5}{11}+\dfrac{5}{11}\right)-\dfrac{1}{3}=\dfrac{-7}{12}\cdot\dfrac{10}{11}-\dfrac{1}{3}=-\dfrac{19}{22}\)
#\(N\)
`1, -3/7 +5/13 - 4/7 +-8/13 = (-3/7 + -4/7) + (5/13 --8/13) = -1+1=0`
`2, -5/14 - 2/14 + 1/8 + 1/8 = (-5/14 - 2/14) + (1/8 + 1/8) = -1/2 + 1/4 =-1/4`
`3, -5/22 -1 +3/2-6/22 = (-5/22 - 6/22) -1+3/2 = -1/2 - 1+3/2 = -3/2 + 3/2 = 0 `
`4, -1/2 + 1/4 + 2/8 -11/77 = -1/2+1/4 +1/4 -11/77 = -1/2 +(1/4+1/4)-1/7 =-1/2 + 1/2 - 1/7 =0-1/7 = -1/7`
`5, 2/5 + 5/4+(-2/5+5/4)-5/2 = 2/5+5/4+-2/5+5/4-5/2 = (2/5+-2/5)+(5/4+5/4)-5/2 = 0+5/2-5/2 = 5/2 - 5/2 = 0`
1: =-3/7-4/7+5/13+8/13
=-1+1=0
2: =-5/14-2/14+1/4
=-1/2+1/4
=-1/4
3: =-5/22-6/22+3/2-1
=-1/2-1+3/2
=0
4: =-4/8+2/8+2/8-1/7=-1/7
5: =2/5+5/4-2/5+5/4-5/2
=5/2-5/2
=0
\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)
Ta có :\(\frac{-5}{12}=\frac{-25}{60}\) \(\frac{7}{-10}=\frac{-42}{60}\)
Vì \(\frac{-25}{60}>\frac{-42}{60}\)
Nên \(\frac{-5}{12}>\frac{7}{-10}\)
\(5A=5^2+5^3+5^4+...+5^{2017}\)
\(A=\frac{5A-A}{4}=\frac{5^{2017}-5}{4}=\frac{5\left(5^{2016}-1\right)}{4}\)
\(\Rightarrow4.A+5=\frac{4.5\left(5^{2016}-1\right)}{4}+5=5\left(5^{2016}-1\right)+5=5^x\)
\(\Rightarrow\frac{5\left(5^{2016}-1\right)}{5}+\frac{5}{5}=\frac{5^x}{5}\Rightarrow5^{2016}-1+1=5^{x-1}\)
\(\Rightarrow5^{2016}=5^{x-1}\Rightarrow x-1=2016\Rightarrow x=2017\)