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31 tháng 12 2020

= 6 - 1 + \(\dfrac{1}{3}\) =\(\dfrac{16}{3}\)

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`

31 tháng 10 2023

a) 2021 - (1/3)² . 3²

= 2021 - 1/9 . 9

= 2021 - 1

= 2020

b) 5/10 + 9 . (-3/2)

= 1/2 - 27/2

= -26/2

= -13

c) -10 . (-2021/2022)⁰ + (2/5)² : 2

= -10 . 1 + 4/25 . 2

= -10 + 8/25

= -68/7

31 tháng 10 2023

\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)

`#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`

Ta có: \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-99}{100}\)

\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)

\(=-\dfrac{10+1}{2\cdot10}=\dfrac{-11}{20}\)

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.

15 tháng 12 2021

\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

15 tháng 12 2021

thanks nhìu

8 tháng 9 2023

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

7 tháng 7 2021

\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{1}{4}\)

\(=\dfrac{9}{12}-\dfrac{30}{12}-\dfrac{20}{12}+\dfrac{3}{12}\)

\(=\dfrac{-38}{12}=\dfrac{-19}{6}\)

13 tháng 12 2018

\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{1}{2\left(x-3\right)}\)

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)

\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)