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1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)
\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)
b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)
\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)
c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)
\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)
\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)
2) \(-\frac{3}{5}-x=0,75\)
=> \(-\frac{3}{5}-x=\frac{3}{4}\)
=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)
b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)
=> \(x=\frac{2}{5}\)
c) |2x - 4| + 1 = 5
=> |2x - 4| = 4
<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Giúp mình với nha cả nhả :<
Cả nhà làm vài ý thui cx được ạ :<
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(0,\left(3\right)+3\frac{1}{3}+0,\left(31\right)\)
\(=\frac{1}{3}+\frac{10}{3}+\frac{31}{99}\)
\(=\frac{11}{3}+\frac{31}{99}\)
\(=\frac{394}{99}.\)
b) \(\frac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\frac{4}{9}+\frac{1219}{990}-\frac{13}{99}\)
\(=\frac{553}{330}-\frac{13}{99}\)
\(=\frac{139}{90}.\)
Bài 2:
\(0,\left(37\right).x=1\)
\(\Rightarrow\frac{37}{99}.x=1\)
\(\Rightarrow x=1:\frac{37}{99}\)
\(\Rightarrow x=\frac{99}{37}\)
Vậy \(x=\frac{99}{37}.\)
Chúc bạn học tốt!
Phương Nguyễn Mai Bạn thử xem ở đây nhé:
Lý thuyết số thập phân hữu hạn. số thập phân vô hạn tuần ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}\left(1-3-\dfrac{1}{2}\right)\)
\(=\dfrac{-247}{110}+\dfrac{1}{3}\cdot\dfrac{-5}{2}=\dfrac{-508}{165}\)
b: \(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}=\dfrac{139}{90}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(0,\left(3\right)+3\dfrac{1}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{19}{45}\)
\(=\dfrac{184}{45}\)
b) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{1789}{990}\)
Bài 2:
a) \(0,\left(37\right)x=1\)
\(\Leftrightarrow\dfrac{37}{99}.x=1\)
\(\Leftrightarrow x=1:\dfrac{37}{99}\)
\(\Leftrightarrow x=\dfrac{99}{37}\)
b) \(0,\left(26\right)x=1,2\left(31\right)\)
\(\Leftrightarrow\dfrac{26}{99}x=\dfrac{1219}{990}\)
\(\Leftrightarrow x=\dfrac{1219}{990}:\dfrac{26}{99}\)
\(\Leftrightarrow x=\dfrac{1219}{260}\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(0,\left(3\right)+\dfrac{10}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{4}{9}\)
\(=\dfrac{33}{9}+\dfrac{4}{9}=\dfrac{37}{9}\)
b: Ta có: \(\dfrac{4}{9}+1.2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{440}{990}+\dfrac{1219}{990}-\dfrac{130}{990}\)
\(=\dfrac{139}{90}\)
c: Ta có: \(2,\left(4\right)\cdot\dfrac{3}{11}\)
\(=\dfrac{22}{9}\cdot\dfrac{3}{11}\)
\(=\dfrac{2}{3}\)
d: Ta có: \(-0,\left(3\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
=0