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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cdot DuyNam\)
\(A=-\dfrac{7}{21}+\left(1+\dfrac{1}{3}\right)\)
\(A=-\dfrac{7}{21}+\dfrac{4}{3}\)
\(A=1\)
\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+-\dfrac{6}{9}\right)\)
\(B=\dfrac{2}{15}+-\dfrac{1}{9}\)
\(B=\dfrac{1}{45}\)
\(C=\left(-\dfrac{1}{5}+\dfrac{3}{12}\right)+-\dfrac{3}{4}\)
\(C=\dfrac{1}{20}+-\dfrac{3}{4}\)
\(C=-\dfrac{7}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)=1/2 . 8/15 - 3/4.47/9
=4/15 - 47/12
=-73/20
b)=2-1/3 . -21/20
=2+7/20
=47/20
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1: =(1+2+...+10)x4=220
Câu 2: Vẫn chưa hiểu mấy cái điểm số trên để làm gì nhưng đáp án là bội số của 3.
không thích thì bỏ qua câu này. :v
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{10}\right)\) . x = \(\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{9}{1}\)
Biến đổi vế phải
\(\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{9}{1}\)
= \(\left(\dfrac{1}{9}+1\right)+\left(\dfrac{2}{8}+1\right)+\left(\dfrac{3}{7}+1\right)+...+\left(\dfrac{8}{2}+1\right)+1\)
= \(\dfrac{10}{10}+\dfrac{10}{9}+\dfrac{10}{8}+\dfrac{10}{7}+...+\dfrac{10}{2}\)
= \(10\left(\dfrac{1}{10}+\dfrac{1}{9}+\dfrac{1}{8}+...+\dfrac{1}{2}\right)\)
⇒ x = 10
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{10}\right)\cdot x=\dfrac{1}{9}+\dfrac{2}{8}+...+\dfrac{9}{1}\)
=>\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{10}\right)\cdot x=\left(\dfrac{1}{9}+1\right)+\left(\dfrac{2}{8}+1\right)+...+\left(\dfrac{8}{2}+1\right)+1\)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)=\dfrac{10}{2}+\dfrac{10}{3}+...+\dfrac{10}{10}\)
=>\(x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)=10\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
=>x=10