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MC
23 tháng 7 2018
\(1)\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{3}{5}\right)\)
\(=\dfrac{179}{50}-\left(\dfrac{118}{60}+\dfrac{36}{60}\right)\)
\(=\dfrac{179}{50}-\dfrac{77}{30}\)
\(=\dfrac{537}{150}-\dfrac{385}{150}\)
\(=\dfrac{152}{150}=\dfrac{76}{75}=1\dfrac{1}{75}\)
23 tháng 7 2018
Câu 1 : Ta có : 179/50 - ( 59/30 + 3/5 ) .
= 179/50 - ( 59/30 + 18/30 ) .
= 179/50 - 77/30 .
= 537/150 - 385/150.
= 76/75 .
Câu 2 : Ta có : ( 3,5 - 70,84 : 23 + 4 ) - 3,375 . 4/9 : 0,78 .
= ( 3,5 - 3,08 + 4 ) - 27/8 . 4/9 : 39/2 .
= 4,42 - 3/2 . 2/39 .
= 221/50 - 1/13 .
= 2873/650 - 50/650 .
= 2823/50 .
= 56,46 .
TV
1
2 tháng 4 2017
7/9:(2+3/4.x)+5/9=23/27
7/9:(2+3/4.x)=23/27-5/9
7/9:(2+3/4.x)=8/27
2+3/4.x=7/9:8/27
2+3/4.x=21/8
3/4.x=21/8-2
3/4.x=5/8
X=5/8:3/4
X=5/6
11 tháng 2 2017
a. \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{7}{9}:\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Rightarrow\frac{3}{4}x=\frac{21}{8}-2\)
\(\Rightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}:\frac{3}{4}\)
\(\Rightarrow x=\frac{5}{6}\)
b. \(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow\frac{-2}{3}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(\frac{-2}{3}\right)\)
\(\Rightarrow x=\frac{-3}{20}\)
c. \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Rightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Rightarrow\left|x\right|=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
2 tháng 7 2020
a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)
\(=-\frac{19}{8}\)
b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{5}{12}\)
c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)
\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)
\(=-\frac{43}{30}\)
d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)
\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)
\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)
\(=\frac{3}{20}\)
e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)
\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)
f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)
\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)
g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)
\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)
\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)
h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)
\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)
\(=\frac{78149}{6210}\)
i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)
\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)
\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)
\(=\frac{55}{4}\)
28 tháng 5 2020
1) \(\left(\frac{7}{8}-\frac{3}{4}\right).1\frac{1}{3}-\frac{2}{7}.\left(3,5\right)^2\)
= \(\frac{1}{8}.\frac{4}{3}-\frac{2}{7}.\frac{49}{4}\) = \(\frac{-10}{3}\)
2) \(\frac{5}{4}:\frac{1}{2}-\frac{3}{4}\) = \(\frac{7}{4}\)
\(\frac{209}{156}\)