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Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)
b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)
\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)
c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)
d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)
\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)
Câu cuối tương tự
\(=7,05-\left(-7,2+3,5+0,85\right)\\ =7,05-\left(-2,85\right)\\ =9.9\)
\(A=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=2\cdot\dfrac{97}{198}=\dfrac{97}{99}\)
a) 73,5 x 35,64 + 73,5 x 64,36
= 73,5 x ( 35,64 + 64,36 )
= 73,5 x 100
= 7350
b) 6,48 x 11,25 - 6,48 x 0,25 - 6,48
= 6,48 x ( 11,25 - 0,25 - 1 )
= 6,48 x 10
= 64,8
c) 250 x 16 - ( 608 + 396 + 3,5 )
= 250 x 16 - 1007,5
= 4000 - 1007,5
= 2992,5
d) ( 440 + 480 : 12 ) - 120 : 0,25 + 7,8 x 3,5 x 2
= 680 - 120 : 0,25 + 7,8 x 3,5 x 2
= 680 - 480 + 54,6
= 200 + 54,6
= 254,6
~ nhé ! Mơn ạ ~
~ Hok tốt ~