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

\(=\dfrac{-4}{27}-2\cdot\dfrac{1}{4}+\dfrac{3}{2}+1\)

\(=\dfrac{-4}{27}-\dfrac{1}{2}+\dfrac{3}{2}+1\)

\(=\dfrac{-4}{27}+2\)

\(=\dfrac{50}{27}\)

\(4.\left(\dfrac{-1}{3}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{1}{2}\right)+1\) 

\(=4.\dfrac{-1}{27}-2.\dfrac{1}{4}+3.\dfrac{1}{2}+1\) 

\(=\dfrac{-4}{27}-\dfrac{1}{2}+\dfrac{3}{2}+1\) 

\(=\dfrac{50}{27}\)

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`

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

16 tháng 7 2021

`(6-2 4/5)*3 1/8-1 3/5:1/4`

`=(6-14/5)*25/8-8/5*4`

`=16/5*25/8-32/5`

`=10-32/5=18/5`

13 tháng 2 2022

=(30/5-14/5).25/8.4

=16/5.25/2

=8.5/1.1

=40

 

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.

12 tháng 12 2021

\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)

12 tháng 12 2021

Thank!

 

`#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`

11 tháng 3 2022

A=4/63

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

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

1 tháng 7 2023

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

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

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)