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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=\)\(1-\frac{1}{43}\)
\(=\)\(\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(s=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)
\(s=1-\frac{1}{43}=\frac{42}{43}\)
chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
dễ mà
Gọi tổng đó là S. Theo đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}=\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{43.46}\)
\(=\dfrac{3}{1}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+.....+\dfrac{3}{43}-\dfrac{3}{46}=3-\dfrac{3}{46}=\dfrac{135}{46}\)
Học tốt nha e
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/1 -1/4 +1/4 - 1/7 +1/7 ........+1/40 - 1/43
A = 1/1 - 1/43
A = 42/43
A=1 - 1/4 + 1/4 - 1/7 + .... + 1/40 - 1/43
= 1 - 1/43
= 42/43
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)
Tính nhanh:
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
Tìm x:
a) \(x+30\%x=-1,3\)
\(\Leftrightarrow x\left(1+30\%\right)=-1,3\)
\(\Leftrightarrow x\left(1+\frac{3}{10}\right)=\frac{-13}{10}\)
\(\Leftrightarrow x.\frac{13}{10}=\frac{-13}{10}\)
\(\Leftrightarrow x=\frac{-13}{10}:\frac{13}{10}=\frac{-13}{10}.\frac{10}{13}=-1\)
Vậy \(x=-1\)
b) \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)