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18 tháng 4 2021

B=\(\dfrac{1}{2}:\left(-1\dfrac{1}{2}\right):1\dfrac{1}{3}:....:\left(-1\dfrac{1}{100}\right)\)

   =\(\dfrac{1}{2}:\dfrac{-3}{2}:\dfrac{4}{3}:....:\dfrac{-101}{100}\)

   =\(\dfrac{1}{2}.\dfrac{-2}{3}.\dfrac{3}{4}........\dfrac{-100}{101}\)

   =\(\dfrac{1.\left(-2\right).3......\left(-100\right)}{2.3.4...........101}\)

   =\(\dfrac{1}{101}\)

4 tháng 10 2021

\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)

4 tháng 10 2021

\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)

\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)

               \(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)

\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)

\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)

19 tháng 9 2023

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right)\)

\(B=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)....\left(\dfrac{1}{2020^2}-\dfrac{2020^2}{2020^2}\right)\)

\(B=\left(\dfrac{1-2^2}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)...\left(\dfrac{1-2020^2}{2020^2}\right)\)

\(B=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}\cdot\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}....\cdot\dfrac{\left(2020-1\right)\left(2020+1\right)}{2020^2}\) 

\(B=\dfrac{-1\cdot3}{2^2}\cdot\dfrac{-2\cdot4}{3^2}\cdot\dfrac{-3\cdot5}{4^2}\cdot....\cdot\dfrac{-2019\cdot2021}{2020}\)

\(B=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-2019}{2\cdot3\cdot4\cdot....\cdot2020}\)

\(B=\dfrac{-1\cdot-1\cdot-1\cdot....\cdot-1}{1}\)

\(B=-1\) (2019 số -1) 

Mà: \(-1< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\)

19 tháng 9 2023

 \(\dfrac{1}{2^2}\)\(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 1 ⇒ 0 > \(\dfrac{1}{2^2}\) - 1 > \(\dfrac{1}{3^2}\) - 1 >..> \(\dfrac{1}{2020^2}\) - 1

Xét dãy số 2; 3; 4;...; 2020 dãy số này có số số hạng là:

        (2020 - 2):1 + 1 = 2019 (số hạng)

Vậy B là tích của 2019 số âm nên B < 0 ⇒ B < \(\dfrac{1}{2}\)

 

 

 

 

 

24 tháng 9 2017

Bài 2:

\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-2}{5}\right)^2\)

\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3.\left(\dfrac{2}{5}\right)^2\)

\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^5\)

\(\dfrac{2}{5}\ne\pm1;\dfrac{2}{5}\ne0\) nên \(x>5\)

Vậy \(x>5\) thoả mãn yêu cầu đề bài.

Chúc bạn học tốt!!!

24 tháng 9 2017

Bài 1:

\(C=\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\)

\(C=\left(\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right).....\left(\dfrac{1}{\left(100-1\right)\left(100+1\right)}\right)\)

\(C=\dfrac{1}{1.3}\dfrac{1}{2.4}.....\dfrac{1}{99.101}=\dfrac{1}{101!}\)

Chúc bạn học tốt!!!

4 tháng 10 2021

ừ bài nâng cao mà bạn ơi :)))

4 tháng 10 2021

\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)

\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)

6 tháng 12 2017

B = .................

Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0

\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)

6 tháng 12 2017

Mình làm câu 1,2 trước, câu 3 sau

Câu 1:

\(\sqrt{x^2}=0\)

=> \(\left(\sqrt{x^2}\right)^2=0^2\)

\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Câu 2:

\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)

\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)

6 tháng 10 2023

A=\(\dfrac{1\cdot4}{2\cdot3}\) \(\cdot\dfrac{2\cdot5}{3\cdot4}\) ...\(\dfrac{2015\cdot2018}{2016\cdot2017}\)

A=\(\dfrac{1\cdot2\cdot3\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2016}\) \(\cdot\dfrac{4\cdot5\cdot...\cdot2018}{3.4\cdot...\cdot2017}\)

A=\(\dfrac{1}{2016}\) \(\cdot\dfrac{2018}{3}\) =\(\dfrac{1009}{336}\)

 

 

20 tháng 12 2017

1/\(\dfrac{-2}{5}\)

2/-21

21 tháng 12 2017

1/ \(4\left(\dfrac{-1}{2}\right)^3+\dfrac{1}{2}:5\)

\(=4.\dfrac{-1}{8}+\dfrac{1}{2}.\dfrac{1}{5}\)

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

\(=\dfrac{-5}{10}+\dfrac{1}{10}\)

\(=\dfrac{-4}{10}\)

\(=\dfrac{-2}{5}\)

2/ \(17\dfrac{1}{5}:\left(-\dfrac{5}{7}\right)-2\dfrac{1}{5}.\left(-\dfrac{7}{5}\right)\)

\(=\dfrac{86}{5}.\left(\dfrac{-7}{5}\right)-\dfrac{11}{5}.\left(\dfrac{-7}{5}\right)\)

\(=\dfrac{-7}{5}.\left(\dfrac{86}{5}-\dfrac{11}{5}\right)\)

\(=\dfrac{-7}{5}.15\)

\(=-21\)

29 tháng 8 2023

\(...A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)....\left(-\dfrac{1998}{1999}\right).\)

Số dấu trừ là : \(\left(1998-1\right):1+1=1998\) là số chẵn

\(\Rightarrow A=\dfrac{1.2.3...1998}{2.3.4...1999}\)

\(\Rightarrow A=\dfrac{1}{1999}\)

29 tháng 8 2023

gợi ý nè

tính hết mấy cái hiệu trong ngoặc rồi nhân lại

vì kết thúc ở số 1999

nên sẽ có 1999 dấu -

nên kq là âm

nhân ra rồi triệt tiêu đi