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\(B=\dfrac{12}{17}.\dfrac{5}{7}-\dfrac{-12}{17}.\dfrac{1}{7}+\dfrac{1}{17.7}.12\\ =\dfrac{12}{17}.\dfrac{5}{7}+\dfrac{12}{17}.\dfrac{1}{7}+\dfrac{12}{17}.\dfrac{1}{7}\\ =\dfrac{12}{17}\left(\dfrac{5}{7}+\dfrac{1}{7}+\dfrac{1}{7}\right)\\ =\dfrac{15}{17}.1=\dfrac{15}{17}\)
\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
a) \(=\left(\left(-\frac{1}{4}-\frac{5}{3}\right)+\frac{7}{33}\right)-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\left(-\frac{23}{12}+\frac{7}{33}\right)+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(-\frac{23}{12}+\frac{15}{12}\right)+\left(\frac{9}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=-\frac{2}{3}-\frac{3}{11}+\frac{48}{49}\)
\(=\frac{65}{1617}\)
b) \(=\frac{11}{125}+\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)
a/ -7/2+3/4-17/12=-42/12+9/12-17/12=-50/12=-25/6
b/ 3/4+[-9/5]-4/3=45/60-108/60-80/60=-142/60=-71/30
c/ -1/12-[21/3-1/3]=-1/12-20/3=-1/12-80/12=-81/12=-27/4
d/ 22/7.[-7/11]+18= -2+18=16
e/ [3/4+2/3]:17/4-3/4=[9/12+8/12]:17/4-3/4=17/12:17/14-3/4=14/12-3/4=14/12-9/12=5/12
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{12}{17}\right)-\frac{7}{5}\)
\(=0+0+0-\frac{7}{5}\)
\(=-\frac{7}{5}\)
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{-12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{-12}{17}\right)-\frac{7}{5}\)
\(=0+0+\frac{24}{17}-\frac{7}{5}\)
\(=\frac{24}{17}-\frac{7}{5}\)
\(=\frac{120}{85}-\frac{119}{85}=\frac{1}{85}\)
a)\(-\frac{6}{7}< \frac{3}{7}< \frac{18}{7}\)
b) \(\frac{17}{35}>\frac{17}{53}\)
c) \(\frac{17}{35}< \frac{17}{-35}\)
d) Ta có : \(\frac{12}{7}=\frac{60}{35}\)
\(\frac{17}{5}=\frac{85}{35}\)
\(\Rightarrow\frac{12}{7}< \frac{17}{5}\)
a)\(\frac{3}{7}\)>\(\frac{18}{7}\)>\(\frac{-6}{7}\)
b)\(\frac{17}{35}\)>\(\frac{17}{53}\)
c)\(\frac{17}{35}\)<\(\frac{17}{-35}\)
d)\(\frac{12}{7}\)<\(\frac{17}{5}\)
hok tốt
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(\dfrac{12}{17}.\dfrac{5}{7}-\left(-\dfrac{12}{17}\right).\dfrac{1}{7}+\dfrac{1}{17+7}\)
\(=\dfrac{12}{17}.\dfrac{5}{7}+\dfrac{12}{17}.\dfrac{1}{7}+\dfrac{1}{24}\)
\(=\dfrac{12}{17}\left(\dfrac{5}{7}+\dfrac{1}{7}\right)+\dfrac{1}{24}\)
\(=\dfrac{12}{17}.\dfrac{6}{7}+\dfrac{1}{24}\)
\(=\dfrac{72}{119}+\dfrac{1}{24}=\dfrac{1847}{2856}\)
Thank you! :)