1 + 3 + 5 + ......+ 997
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![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2x-1\right)\)
\(-\frac{2}{3}x+\frac{1}{6}=\frac{2}{3}x-\frac{1}{3}\)
\(-\frac{2}{3}x-\frac{2}{3}x=-\frac{1}{3}-\frac{1}{6}\)
\(-\frac{4}{3}x=-\frac{1}{2}\)
\(x=\left(-\frac{1}{2}\right):\left(-\frac{4}{3}\right)\)
\(x=\frac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- \(\dfrac{2}{3}\).(\(x-\dfrac{1}{4}\)) = \(\dfrac{1}{3}\)(2\(x-1\))
- \(\dfrac{2}{3}x\) + \(\dfrac{1}{6}\) = \(\dfrac{2}{3}x\) - \(\dfrac{1}{3}\)
\(\dfrac{2}{3}x+\dfrac{2}{3}x\) = \(\dfrac{1}{6}+\dfrac{1}{3}\)
\(\dfrac{4}{3}x=\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{1}{2}\) : \(\dfrac{4}{3}\)
\(x\) = \(\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mik nghĩ môn toán thì sẽ khó hơn vì nó lí luận dài dòng phức tạp nhưng cả hai môn toán và anh đều có tầm quan trọng như nhau
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a;a+1;a+2;a+3 là bốn số nguyên liên tiếp
nên \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)⋮4!\)
=>\(A⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\dfrac{5}{2}-x=\dfrac{27}{8}\)
=>\(x=\dfrac{5}{2}-\dfrac{27}{8}=\dfrac{20}{8}-\dfrac{27}{8}=-\dfrac{7}{8}\)
b: \(-\dfrac{1}{3}+\dfrac{4}{10}x=0,2\)
=>\(\dfrac{2}{5}x=0,2+\dfrac{1}{3}=\dfrac{1}{5}+\dfrac{1}{3}=\dfrac{8}{15}\)
=>\(x=\dfrac{8}{15}:\dfrac{2}{5}=\dfrac{8}{15}\cdot\dfrac{5}{2}=\dfrac{40}{30}=\dfrac{4}{3}\)
c: \(\left(3x+2\right)^2=\dfrac{25}{49}\)
=>\(\left[{}\begin{matrix}3x+2=\dfrac{5}{7}\\3x+2=-\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{5}{7}-2=-\dfrac{9}{7}\\3x=-\dfrac{5}{7}-2=-\dfrac{19}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{3}{7}\\x=-\dfrac{19}{21}\end{matrix}\right.\)
Bài 1:
a: \(\dfrac{5}{6}+\dfrac{-11}{30}+\dfrac{7}{5}\)
\(=\dfrac{25}{30}-\dfrac{11}{30}+\dfrac{42}{30}=\dfrac{56}{30}=\dfrac{28}{15}\)
b: \(\left(-\dfrac{4}{5}\right)^2\cdot\dfrac{57}{2}+\dfrac{-7}{2}\cdot\left(-\dfrac{4}{5}\right)^2\)
\(=\left(-\dfrac{4}{5}\right)^2\left(\dfrac{57}{2}-\dfrac{7}{2}\right)=\dfrac{16}{25}\cdot25=16\)
c: \(\left(\dfrac{7}{25}-1\dfrac{2}{9}\right)-\left(\dfrac{23}{54}-\dfrac{18}{25}\right)+\dfrac{-31}{54}\)
\(=\dfrac{7}{25}-\dfrac{11}{9}-\dfrac{23}{54}+\dfrac{18}{25}-\dfrac{31}{54}\)
\(=\left(\dfrac{7}{25}+\dfrac{18}{25}\right)-\left(\dfrac{23}{54}+\dfrac{31}{54}\right)-\dfrac{11}{9}=1-1-\dfrac{11}{9}=-\dfrac{11}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: \(x-\dfrac{1}{3}=\dfrac{5}{2}\)
=>\(x=\dfrac{5}{2}+\dfrac{1}{3}=\dfrac{15}{6}+\dfrac{2}{6}=\dfrac{17}{6}\)
b: \(\left(1-2x\right)^2=\dfrac{4}{9}\)
=>\(\left(2x-1\right)^2=\dfrac{4}{9}\)
=>\(\left[{}\begin{matrix}2x-1=\dfrac{2}{3}\\2x-1=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{3}\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Câu 1:
a: \(-\dfrac{4}{3}+\dfrac{3}{2}:\dfrac{9}{4}\)
\(=\dfrac{-4}{3}+\dfrac{3}{2}\cdot\dfrac{4}{9}\)
\(=\dfrac{-4}{3}+\dfrac{2}{3}=-\dfrac{2}{3}\)
b: \(\left(\dfrac{10}{9}+\dfrac{13}{7}\right)-\left(\dfrac{19}{9}-\dfrac{1}{7}\right)\)
\(=\dfrac{10}{9}+\dfrac{13}{7}-\dfrac{19}{9}+\dfrac{1}{7}=\left(\dfrac{10}{9}-\dfrac{19}{9}\right)+\left(\dfrac{13}{7}+\dfrac{1}{7}\right)\)
=-1+2
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
1/5 giờ=12 phút
\(1\dfrac{1}{3}phút=80\left(giây\right)\)
7m45cm=745cm
Ta có:
1 + 3 + 5 + ... + 997
Tổng trên có số số hạng là:
( 997 - 1 ) : 2 + 1 = 499 ( số hạng )
Giá trị của tổng trên là:
( 997 +1 ) x 499 : 2 = 249001
Vậy giá trị của tổng trên là 249001.
Tổng trên có số số hạng là: (997-1):2+1 = 499 (số)
⇒ 1+3+5+...+997 = (997+1)x499:2=249001
Vậy 1+3+5+...+997=249001