Tính:
\(\left(1+\frac{7}{9}\right)\times\left(1+\frac{7}{20}\right)\times\left(1+\frac{7}{33}\right)\times\left(1+\frac{7}{48}\right)\times....\times\left(1+\frac{7}{180}\right)\)
Thần đòng nào giỏi toán vào giúp mik zới!Chiều nay thi rồi huhu!!!!!!!!!
(1+7/9).(1+7/20).(1+7/33).(1+7/48)......(1+7/180)
=16/9.27/20.40/33.55/48........187/180
=2.8/1.9 . 3.9/2.10 . 4.10/3.11 . 5.11/4.12 ........ 11.17/18.10
=(2.3.4.5.......11).(8.9.10......17)/(1.2.3.4.....18).(9.10.11.12......18)
=11.8/1.18=88/18=44/9
Ta có: (1 + 7/9) x (1 + 7/20) x (1 + 7/33) x (1 + 7/48) x ...x (1 + 7/180)
= (9/9 + 7/9) x (20/20 + 7/20) x (33/33 + 7/33) x (48/48 + 7/48) x ...x (180/180 + 7/180)
= 16/9 x 27/20 x 40/33 x 55/48 x ......x 187/180
=
tới đây tự làm tiếp nha! Bí rồi! Chỉ gợi ý phần đầu thôi! Các bạn khác có gì lấy phần đầu bài của mình sau đó làm tiếp nha!
A = \(\dfrac{-19}{9}\times\dfrac{1}{2}-\dfrac{4}{11}\times\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)\)
B = \(\left(-\dfrac{15}{6}\right)\div\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}\times\dfrac{-11}{2}\)
C = \(\dfrac{3}{4}\times\left(-8\right)-\dfrac{1}{3}\times\dfrac{-7}{2}-\dfrac{5}{18}\)
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
Tính hợp lí: \(\left(\dfrac{3}{7}\right)^5\times\left(\dfrac{7}{3}\right)^{-1}\times\left(\dfrac{5}{3}\right)^6\div\left(\dfrac{343}{625}\right)^{-2}\)
\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{5^6}{7^6}:\dfrac{5^8}{7^6}=\dfrac{1}{25}\)
1) A = \(\left(-\dfrac{25}{27}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\)
2) B = \(\dfrac{10\dfrac{3}{10}-\left(9,5-0,25\times18\right)\div0,5}{1\dfrac{1}{5}-1\dfrac{1}{2}}\)
3) C = \(\dfrac{3}{49}\times\dfrac{19}{2}-\dfrac{3}{49}\times\dfrac{5}{2}-\left(\dfrac{1}{20}-\dfrac{1}{4}\right)^2\times\left(\dfrac{-1}{2}-\dfrac{193}{14}\right)\)
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
a)\(\left(\dfrac{17}{10}+7-8,7\right)\div\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\times\left(12,98\times0,25\right)+12,5\)
b)\(1\dfrac{2}{24}\times5\dfrac{2}{5}\times2\times3\dfrac{7}{9}\times2\times\dfrac{2}{17}\)
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
\(\dfrac{9}{48}\times\left(-2,4\right)+\left(\dfrac{1}{4}+\dfrac{13}{20}\right)\div2\)
9/48.(-2,4)+9/10:2
=-9/20+9/10:2
=-9/20+9/20
=0
Tìm x
a, \(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
b, \(\left(x-\dfrac{3}{2}\right)\times\left(2\times x-1\right)>0\)
c, \(\left(x-\dfrac{4}{7}\right)\div\left(x+\dfrac{1}{2}\right)>0\)
d, \(\left(2\times x-5\right)\div\left(x+1\dfrac{3}{4}\right)< 0\)
Giúp mk với
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
\(F=\dfrac{1}{2\times\left(1+2\right)}+\dfrac{1}{2\times\left(1+2+3\right)}+....+\dfrac{1}{2\times\left(1+2+3+...+9\right)}\)
Tính \(\left(1+\dfrac{1}{1+2}\right)\times\left(1+\dfrac{1}{1+2+3}\right)\times\left(1+\dfrac{1}{1+2+3+4}\right)\times...\times\left(1+\dfrac{1}{1+2+3+...+997}\right)\)
