Tính M=\(\dfrac{5^3}{1.6}+\dfrac{5^3}{6.11}+...+\dfrac{5^3}{26.31}\)
Giúp mk nha!
Tính giá trị của biểu thức: \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
`A = ( 5^2 )/( 1*6)+(5^2)/(6*11)+.....+(5^2)/(26*31)`
`= 5*( 5/( 1*6)+ 5/(6*11)+.....+5/(26*31))`
`= 5*( 1 - 1/6 + 1/6 - 1/11 +....+1/26 - 1/31 )`
`= 5*( 1 - 1/31 )`
`= 5 * 30/31 = 150/31`
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
A=\(\dfrac{5^{2}}{1.6}+\dfrac{5^{2}}{6.11}+...+\dfrac{5^{2}}{26.31}\)
Chứng tỏ A<5
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+..+\dfrac{5}{26.31}\right)\)
\(A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(A=5\left(1-\dfrac{1}{31}\right)\)
\(A=5-\dfrac{1}{155}\)
\(A< 5\rightarrowđpcm\)
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+.......\dfrac{5^2}{26.31}\)
\(\Leftrightarrow A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+.........+\dfrac{5}{26.31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+..........+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5.\dfrac{30}{31}=\dfrac{150}{31}\)
Bài 1 : Tính
a) A = \(\left(\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(\dfrac{55}{123}+\dfrac{555}{1234}+\dfrac{5555}{12345}\right)\)
b) B = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{101.106}\)
c) C = \(\dfrac{2x^2+3x-1}{3x-2}\) với \(\left|x-1\right|=2\)
a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
tìm X
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}\)
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}=1\)
\(=x\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\right)=1\)
\(=x\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\right)=1\)
\(=x\left(1-\dfrac{1}{31}\right)=1\)
\(\Rightarrow x=1:\left(1-\dfrac{1}{31}\right)=\dfrac{31}{30}\)
1/Tính:
a. S=\(\dfrac{5^2}{1.6}\) + \(\dfrac{5^2}{6.11}\)+ \(\dfrac{5^2}{11.16}\) + \(\dfrac{5^2}{16.21}\) + \(\dfrac{5^2}{21.26}\)
b. (1 - \(\dfrac{1}{2}\)) . (1 - \(\dfrac{1}{3}\) ) . (1- \(\dfrac{1}{4}\) ) . ( 1 - \(\dfrac{1}{5}\) ) .... ( 1 - \(\dfrac{1}{19}\) ) . ( 1 - \(\dfrac{1}{20}\))
Mk cần gấp lắm ~help me please~
Giải:
a) S=52/1.6+52/6.11+52/11.16+52/16.21+52/21.26
S=5.(5.1/6+5/6.11+5/11.16+5/16.21+5/21.26)
S=5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26)
S=5.(1/1-1/26)
S=5.25/26
S=125/26
b) (1-1/2).(1-1/3).(1-1/4).(1-1/5).....(1-1/19).(1-1/20)
=1/2.2/3.3/4.4/5.....18/19.19/20
=1.2.3.4.....18.19/2.3.4.5.....19.20
=1/20
Chúc bạn học tốt!
Tính giá trị các biểu thức:
A=(-1)+(-5)+(-9)+...+(-101)
B=\(\dfrac{-5}{17}\).\(\dfrac{8}{19}\)+\(\dfrac{-12}{17}\). \(\dfrac{8}{19}\) - \(\dfrac{11}{19}\)
C=\(\dfrac{10}{1.6}\)+\(\dfrac{10}{6.11}\)+\(\dfrac{10}{11.16}\)+...+\(\dfrac{10}{2016.2021}\)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
Tính nhanh:
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+....+\dfrac{5}{96.101}\)
Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}-\dfrac{1}{101}\)
\(=\dfrac{101-1}{101}\)
\(=\dfrac{100}{101}\)
[Bài này nhiều câu lắm, những 12 câu cơ, sợ các bạn thấy dài là nản! Tí nữa đăng tiếp! - 03/12 câu]
Tính tổng sau bằng phương pháp hợp lý nhất:
1) A = \(\dfrac{5^2}{1.6}\) + \(\dfrac{5^2}{6.11}\) + ... + \(\dfrac{5^2}{26.31}\)
2) B = \(\dfrac{6}{15.18}\) + \(\dfrac{6}{18.21}\) + \(\dfrac{6}{21.24}\) + ... + \(\dfrac{6}{87.90}\)
3) C = \(\dfrac{3^2}{8.11}\) + \(\dfrac{3^2}{11.14}\) + ... + \(\dfrac{3^2}{197.200}\)
Các bạn cố gắng giải đầy đủ giúp mình!
a, \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(A=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(A=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
b, \(B=\dfrac{6}{15.18}+\dfrac{6}{18.21}+...+\dfrac{6}{87.90}\)
\(B=2\left(\dfrac{3}{15.18}+\dfrac{3}{18.21}+...+\dfrac{13}{87.90}\right)\)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)=2.\dfrac{1}{18}=\dfrac{1}{9}\)
c, \(C=\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
\(C=3\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)=3.\dfrac{3}{35}=\dfrac{9}{35}\)
Chúc bạn học tốt!!!
Điền dấu thích hợp (>,<,=) vào chỗ trống.
a,\(\dfrac{-3}{5}+\dfrac{-11}{5}...\dfrac{-7}{3}+\dfrac{4}{7}\)
b,\(\dfrac{-7}{11}+\dfrac{-15}{11}...\dfrac{-4}{3}+\dfrac{-7}{5}\)
Giúp mk nha :>>. Mk cảm ơn.