Ta có : \(\dfrac{1.3.5+2.6.10+4.12.20 +7.21.35 }{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+1.2.3.2.5.2+1.4.3.4.5.4+1.7.3.7.5.7}{1.5.7+1.2.5.2.7.2+1.4.5.4.7.4+1.7.5.7.7.7}\)

\(=\dfrac{1.\left(1.3.5\right)+2.\left(1.3.5\right)+4.\left(1.3.5\right)+7.\left(1.3.5\right)}{1.\left(1.5.7\right)+2.\left(1.5.7\right)+4.\left(1.5.7\right)+7.\left(1.5.7\right)}\)

\(=\dfrac{1.3.5.\left(1+2+4+7\right)}{1.5.7.\left(1+2+4+7\right)}\)

\(=\dfrac{3}{7}\)

Chứng tỏ : \(8^7-2^{18}\) chia hết cho 14

\(Ta \) \(có\) : \(8^7-2^{18}\)

\(=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}\)

\(=2^{18}.2^3-2^{18}\)

\(=2^{18}.\left(2^3-1\right)\)

\(=2^{18}.\left(8-1\right)\)

\(=2^{18}.7\)

\(=2^{17}.2.7\)

\(=2^{17}.14\) ( chia hết cho 14 )

\(\Rightarrow8^7-2^{18}\) chia hết cho 14

ĐỀ NÈ BẠN :

Bài 1 : Thực hiện phép tính :

a) A=1.2.3...9-1.2.3...8-1.2.3...8.8

b) B=\(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

c) \(70\left(\dfrac{131313}{565656}+\dfrac{131313}{727272}+\dfrac{131313}{909090}\right)\)

d) B=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{64.69}\)

Có bạn gửi cho em và trong đó nói nếu em không gửi cho người khác thì kì thi sắp tới em sẽ được điểm kém nữa cơ. Nhưng e không gửi cho ai cả vì nghĩ như vậy là không đúng .

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=1-\dfrac{1}{50}\)

\(A=\dfrac{49}{50}\)

ở đây học toán chứ không phải tiếng anh

Bài 5: Giải:

Xe máy đã đi được :

\(108.\dfrac{5}{6}=90\left(km\right)\)

Xe máy còn cách Huế :

\(108-90=18\left(km\right)\)

Đáp số : \(18km\)

Bài 4 :

\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{97.99}\)

\(=2\left(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{97.99}\right)\)

\(=3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)

\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=3\left(1-\dfrac{1}{99}\right)\)

\(=3\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(=3.\dfrac{98}{99}\)

\(=\dfrac{3.98}{99}\)

\(=\dfrac{1.98}{33}\)

\(=\dfrac{98}{33}\)

Bài 2:

a) \(\dfrac{2}{3}.x+\dfrac{1}{4}=\dfrac{7}{12}\)

\(\dfrac{2}{3}.x=\dfrac{7}{12}-\dfrac{1}{4}\)

\(\dfrac{2}{3}.x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{2}{3}\)

\(x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}:2.x=-1\)

\(\dfrac{1}{3}:2.x=-1-\dfrac{1}{2}\)

\(\dfrac{1}{3}:2.x=\dfrac{-3}{2}\)

\(2.x=\dfrac{1}{3}:\dfrac{-3}{2}\)

\(2.x=\dfrac{-2}{9}\)

\(x=\dfrac{-2}{9}:2\)

\(x=\dfrac{-1}{9}\)

Vậy \(x=\dfrac{-1}{9}\)

Bài 1 :

a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(=2\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)

\(=2.\dfrac{32}{99}\)

\(=\dfrac{2.32}{99}\)

\(=\dfrac{64}{99}\)

b) \(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)

\(=2\left(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\right)\)

\(=3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=3\left(1-\dfrac{1}{51}\right)\)

\(=3.\dfrac{50}{51}\)

\(=\dfrac{3.50}{51}\)

\(=\dfrac{1.50}{17}\)

\(=\dfrac{50}{17}\)