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a) x + 2/5 = -4/3
x = -4/3 - 2/5
x = -26/15
b) -5/6 + 1/3 x = (-1/2)²
-5/6 + 1/3 x = 1/4
1/3 x = 1/4 + 5/6
1/3 x = 13/12
x = 13/12 : 1/3
x = 13/4
c) 7/12 - (x + 7/6) . 6/5 = (-1/2)³
7/12 - (x + 7/6) . 6/5 = -1/8
(x + 7/6) . 6/5 = 7/12 + 1/8
(x + 7/6) . 6/5 = 17/24
x + 7/6 = 17/24 : 6/5
x + 7/6 = 85/144
x = 85/144 - 7/6
x = -83/144
\(a,x+\dfrac{2}{5}=-\dfrac{4}{3}\\ \Rightarrow x=-\dfrac{26}{15}\\ b,-\dfrac{5}{6}+\dfrac{1}{3}x=\left(-\dfrac{1}{2}\right)^2\\ \Rightarrow-\dfrac{5}{6}+\dfrac{1}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{3}x=\dfrac{13}{12}\\ \Rightarrow x=\dfrac{13}{4}\\ c,\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\left(-\dfrac{1}{2}\right)^3\\ \Rightarrow\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=-\dfrac{1}{8}\\ \Rightarrow\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\dfrac{17}{24}\\ \Rightarrow x+\dfrac{7}{6}=\dfrac{85}{144}\\ \Rightarrow x=-\dfrac{83}{144}.\)
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
a) 7/2 - (3/4 + 1/5)
= 7/2 - 19/20
= 51/20
b) 12/23 . 7/13 + 11/23 . 7/13
= 7/13 . (12/23 + 11/23)
= 7/13 . 1
= 7/13
c) |-2| - (5/9 - 2/3)² : 4/27
= 2 - 1/81 : 4/27
= 2 - 1/12
= 23/12
a) \(\dfrac{4}{9}+\dfrac{1}{4}=\dfrac{25}{36}\)
b) \(\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}\right)+\dfrac{1}{3}\cdot\left(-\dfrac{1}{5}\right)=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)=\dfrac{1}{3}\cdot-1=-\dfrac{1}{3}\)
c) \(\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(1-\dfrac{1}{2}\right)^2\right]=\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{5}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{1}{5}-0=\dfrac{1}{5}\)
`#3107.101107`
`a)`
\(\dfrac{4}{9}+\dfrac{1}{4}=\dfrac{16}{36}+\dfrac{9}{36}=\dfrac{25}{36}\)
`b)`
\(\dfrac{1}{3}\cdot\left(\dfrac{-4}{5}\right)+\dfrac{1}{3}\cdot\left(-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{3}\cdot\left(-1\right)\)
\(=-\dfrac{1}{3}\)
`c)`
\(\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(1-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{1}{5}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=\dfrac{1}{5}-0\)
\(=\dfrac{1}{5}\)