Thực hiện phép tính :\(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
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2
28 tháng 5 2022
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
30 tháng 3 2021
Ta có: \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-99}{100}\)
\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=-\dfrac{10+1}{2\cdot10}=\dfrac{-11}{20}\)
31 tháng 3 2021
Phải thế này nha bạn!
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}...\dfrac{10^2-1^2}{10^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}...\dfrac{\left(10+1\right)\left(10-1\right)}{10.10}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{\left(10+1\right)\left(10-1\right)}{10.10}\)
\(=\dfrac{\left[1.2.3...\left(10+1\right)\right]\left[3.4.5...\left(10-1\right)\right]}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
\(=\left(10+1\right).\dfrac{1}{2.10}\)
\(=\dfrac{11}{20}\)
Theo mình nghĩ phải như thế này.
LP
15 tháng 12 2021
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
27 tháng 1 2023
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
=1/x-1/x+2014
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
15 tháng 10 2023
`#3107.101107`
`[(1/3)^21 \div (1/3)^3 + 2*(1/9)^9] \div [(1/3)^6]^3`
`= { (1/3)^(21 - 3) + 2*[ (1/3)^2]^9} \div (1/3)^(6*3)`
`= [ (1/3)^18 + 2*(1/3)^18) \div (1/3)^18`
`= [(1/3)^18 * (1 + 2)] \div (1/3)^18`
`= [(1/3)^18 * 3] \div (1/3)^18`
`= (1/3)^18 \div (1/3)^18 * 3`
`= 1 * 3`
`= 3`
TH
4 tháng 12 2018
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
29 tháng 7 2021
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)