Tính giá trị biểu thức sau: B= (2017-\(\dfrac{1}{4}\)-\(\dfrac{2}{5}-\dfrac{3}{6}-\dfrac{4}{7}-...-\dfrac{2017}{2020}\)): (\(\dfrac{1}{20}\) \(\dfrac{1}{25}\) \(\dfrac{1}{30}\) \(\dfrac{1}{35} ... \...

0

LD

Lâm Đỗ

3 tháng 7 2018

Tính giá trị biểu thức sau :

\(B=\left(2017-\dfrac{1}{4}-\dfrac{2}{5}-\dfrac{3}{6}-\dfrac{4}{7}-....-\dfrac{2017}{2020}\right):\left(\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+\dfrac{1}{35}+....+\dfrac{1}{10100}\right)\)

0

TZ

TORO ZANE

26 tháng 7 2017

tính B:

B=\(\left(2017-\dfrac{1}{4}-\dfrac{2}{5}-\dfrac{3}{6}-\dfrac{4}{7}-...-\dfrac{2017}{2020}\right):\left(\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+\dfrac{1}{35}+...+\dfrac{1}{10100}\right)\)là:

Các anh chị giải giùm em ::::::::::::))))))))))))))))))))

1

TZ

TORO ZANE

26 tháng 7 2017

Xuân Tuấn Trịnhvăn tàiNguyen Bao LinĐịnh Quang ( Real )hCold WindKhánh LinhPhương An

Bài 1: Tính giá trị của biểu thức sauA=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0Mong mn giúp...

TN

thanh như

17 tháng 4 2022

1

NL

Nguyễn Lê Phước Thịnh

29 tháng 6 2023

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

Tính bằng cách hợp lí giá trị của biểu thức. A = \(\left(3-\dfrac{1}{4} +\dfrac{3}{2}\right)\)- \(\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)\)-\(\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\) B...

HP

Hồng Phong Đoàn

19 tháng 10 2023

2

NM

Nguyễn Minh Dương

VIP

19 tháng 10 2023

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

Đúng(3)

NP

Ngô Phúc Nhã Uyên

VIP

29 tháng 10 2023

-_-

Đúng(1)

N

nito

13 tháng 8 2023

tìm x e Q

a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

Tính giá trị của biểu thức: a) (-7,05 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\left(-\dfrac{1}{5}\right)\) : (-3) b) \(\left(\dfrac{3}{25}-1,12\right)\) : \(\dfrac{3}{7}\) :...

1

NN

Ng Ngọc

13 tháng 8 2023

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

Đúng(3)

LP

Lê Phương Linh

20 tháng 11 2023

1

BA

Bùi Anh Đức

21 tháng 11 2023

a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)

= \(-\dfrac{73}{10}\) : - 5

= \(\dfrac{73}{50}\)

b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)

= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)

= \(\dfrac{2}{41}\)

LN

Lê Ngọc Anh

17 tháng 3 2022

Tính giá trị của biểu thức sau:

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)

1

DC

dâu cute

17 tháng 3 2022

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

NK

Nguyễn Kiều Anh

18 tháng 9 2021

Tính nhanh:

a, \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

b, \(\left(-\dfrac{1}{4}+\dfrac{7}{35}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)