2x-505= -340
x+304=505
x-505=304
720:(x-17)=12
Bài 1: A=2/3*7 + 2/7*11 + 2/11*15+ ... +2/99*103 Bài 2: A=7/2 + 7/6 + 7/12 + 7/20 + 7/30 + 7/42 + 7/56 + 7/72 + 7/90 Bài 3: A=505/10*1212 + 505/12*1414 + 505/14*1616 +...+ 505/96*9898 Bài 4: A=2/1*3 - 4/3*5 - 6/5*7 - ... - 20/19*21 Bài 5: A=1 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56 - 17/72 + 19/90 :>
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
Bài 1:
Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)
Bài 2:
Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)
\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)
\(=7\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{63}{10}\)
x:2:4+505=579.Tìm x
ta co x :2 : 4 + 505 +579
=> x : 2 : 4 = 579 - 505 = 74
=> x:2 = 74 x 4 = 296
=> x = 296m x 2
= 592
x:2:4+505=579
x:2:4=579-505
x:2=74x4
x=296x2
x=592
x:2:4+505=579
x:2:4=579-505
x:2:4=74
x:4=74*4
x:4=296
x=296*4
x=592
\(\frac{505}{1212}\) +\(\frac{505}{2020}\)+\(\frac{505}{3030}\)+\(\frac{505}{4242}\)+\(\frac{505}{5656}\)
Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
$25/24$ bạn nhé
X+2/505 + x+3/2015= x+6/1006 + x+506/504
505 + 495 = ???
100 x 10 = ???
đáp án:
505 + 495 = 1000
100 . 10 = 1000
chúc bạn học tốt !!!
\(505+495=1000\)
\(100\times10=1000\)
_Chúc bạn học tốt_
( 505 / 707 + 222 / 333 ) . x = 404 / 909
\(\left(\frac{505}{707}+\frac{222}{333}\right)\cdot x=\frac{404}{909}\)
=> \(\left(\frac{5}{7}+\frac{2}{3}\right)\cdot x=\frac{4}{9}\)
=> \(\frac{29}{21}\cdot x=\frac{4}{9}\)
=> \(x=\frac{4}{9}:\frac{29}{21}\)
=> \(x=\frac{28}{87}\)
TK mk nha!
( 505/707+ 222/333) . x = 404/ 909
29/21 . x = 4/9
x= 4/9 : 29/21
x= 4/9 . 21/29
x= 28/87
Vậy x= 28/87
So sánh 2^2018 và 17^505 giúp vs các bạn. Ek
Ta có : 16<17 => 16505<17505
=> (24)505<17505
=> 22020<17505
Mà 22018<22020
=> 22018<17505
Vậy 22018<17505.
\(\frac{230+x}{505+x}=\frac{4}{5}\)
Ta có :\(\frac{230+x}{505+x}=\frac{4}{5}\)
\(\Rightarrow5.\left(230+x\right)=4.\left(505+x\right)\)
\(\Rightarrow1150+5x=2020+4x\)
\(\Rightarrow5x-4x=2020-1150\)
\(\Rightarrow x=870\)
Vậy x =870
Chứng minh 1/505+2/504+3/503+...+504/2+505/1>2025