không quy đồng mẫu hãy so sánh A=-9/10^2010 + -19/10^2011 ; B=-9/10^2011 + -19/10^2010
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( - \(\dfrac{2}{3}\) + 1\(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)) - \(\dfrac{24}{10}\)
= - \(\dfrac{2}{3}\) + \(\dfrac{5}{4}\) - \(\dfrac{1}{6}\) - \(\dfrac{24}{10}\)
= \(\dfrac{-40}{60}\) + \(\dfrac{25}{60}\) - \(\dfrac{10}{60}\) - \(\dfrac{144}{60}\)
= \(-\dfrac{169}{60}\)
`1/5 + 1/6 -1/10 +1/12 + 1/7 -1/14`
`= (1/5-1/10)+(1/6+1/12)+(1/7-1/14)`
`= (2/10-1/10) +(2/12+1/12) +(2/14-1/14)`
`= 1/10 + 1/4 + 1/14`
`= 4/40 + 10/40 +1/14`
`= 14/40 + 1/14`
`=7/20 +1/14`
`=59/140`
\(B=\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{7}-\dfrac{1}{14}\)
\(B=\left(\dfrac{1}{5}-\dfrac{1}{10}\right)+\left(\dfrac{1}{6}+\dfrac{1}{12}\right)+\left(\dfrac{1}{7}-\dfrac{1}{14}\right)\)
\(B=\left(\dfrac{2}{10}-\dfrac{1}{10}\right)+\left(\dfrac{2}{12}+\dfrac{1}{12}\right)+\left(\dfrac{2}{14}-\dfrac{1}{14}\right)\)
\(B=\dfrac{1}{10}+\dfrac{1}{4}+\dfrac{1}{14}\)
\(B=\dfrac{14}{140}+\dfrac{35}{140}+\dfrac{10}{140}\)
\(B=\dfrac{59}{140}\)
Câu `a` chỗ `3'7` là `3/7` nhỉ?
`a)5/11 . 6/11:3/7`
`=30/121 . 7/3`
`=70/121`
`b)5/2+13/4:2-17/4`
`=5/2+13/8-17/4`
`=33/8-17/4`
`=-1/8`
`8/15-x:3/4=11/18`
`x:3/4=8/15-11/18`
`x:3/4=-7/90`
`x=-7/90 . 3/4`
`x=-7/120`
\(\dfrac{8}{15}-x:\dfrac{3}{4}=\dfrac{11}{18}\)
\(x:\dfrac{3}{4}=\dfrac{8}{15}-\dfrac{11}{18}\)
\(x:\dfrac{3}{4}=-\dfrac{7}{90}\)
\(x=-\dfrac{7}{90}.\dfrac{3}{4}\)
\(x=-\dfrac{7}{120}\)
Vậy \(x=-\dfrac{7}{120}\)
\(\left(x-3\right)^{2x}:\left(x-3\right)^x=1\)
\(\Rightarrow\left(x-3\right)^{2^x}:\left(x-3\right)^x=1\)
\(\Rightarrow\left(x-3\right)^2:\left(x-3\right)^x=1\)
Mà kết quả ở đây bằng 1 nên chỉ có 1 phép chia duy nhất là \(1:1=1\)
Vậy \(\left(x-3\right)^2=\left(x-3\right)^x=1;\)
\(\Rightarrow\left(x-3\right)=\left(x-3\right)=1\) vì chỉ có lũy thừa của 1 mới bằng 1.
\(\Rightarrow x=3+1=4\)
\(A=\dfrac{10^{11}+1}{10^{12}-1}\)
\(\Rightarrow10A=\dfrac{10^{11}+1}{10^{12}-1}.10\)
\(\Rightarrow10A=\dfrac{10\left(10^{11}+1\right)}{10^{12}-1}\)
\(\Rightarrow10A=\dfrac{10^{12}-10}{10^{12}-1}\)
\(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow10B=\dfrac{10^{10}+1}{10^{11}+1}.10\)
\(\Rightarrow10B=\dfrac{\left(10^{10}+1\right).10}{10^{11}+1}\)
\(\Rightarrow10B=\dfrac{10^{11}+10}{10^{11}+1}\)
Ta thấy:
\(10^{12}-1>10^{12}-10>0\Rightarrow10A< 1\)
\(0< 10^{11}+1< 10^{11}+10\Rightarrow10B>1\)
Mà \(10A< 1;10B>1\)
\(\Rightarrow B>A\).
\(a.\dfrac{7}{8}+\dfrac{1}{2}\text{=}\dfrac{11}{8}\)
\(b.\dfrac{5}{6}+\left(-2\right)\text{=}\dfrac{-7}{6}\)
\(c.\dfrac{2}{5}+\dfrac{-3}{8}\text{=}\dfrac{1}{40}\)
\(d.\dfrac{5}{7}-\dfrac{3}{8}\text{=}\dfrac{19}{56}\)
\(e.\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{7}{6}\text{=}\dfrac{17}{12}\)
\(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-90}{10^{2011}}+\dfrac{-19}{10^{2011}}=\dfrac{-109}{10^{2011}}\)
\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-190}{10^{2011}}=\dfrac{-199}{10^{2011}}\)
Mà \(\dfrac{-109}{10^{2011}}>\dfrac{-199}{10^{2011}}\)
\(\Rightarrow A>B\).