Tính A, biết: \(A=\left(1 \dfrac{1}{1 2}\right)\left(1 \dfrac{1}{1 2 3}\right)\left(1 \dfrac{1}{1 2 3 4}\right)...\left(1 \dfrac{1}{1 2 3 ... 996}\right)\left(1 \dfrac{1}{1 2 3 ... 997}\right)\)Lưu ý...

LD

Luân Đào

11 tháng 11 2017

Tính A, biết: \(A=\left(1+\dfrac{1}{1+2}\right)\left(1+\dfrac{1}{1+2+3}\right)\left(1+\dfrac{1}{1+2+3+4}\right)...\left(1+\dfrac{1}{1+2+3+...+996}\right)\left(1+\dfrac{1}{1+2+3+...+997}\right)\)Lưu ý giữa các dấu ngoặc là dấu nhân \(\left(.\right)\), không phải dấu cộng. Vì nếu dấu cộng thì dễ rồi. Giúp mình với...

Đọc tiếp

#Toán lớp 7

0

HS

Hà Sỹ Bách

24 tháng 3 2021

Tính \(\left(1+\dfrac{1}{1+2}\right)\times\left(1+\dfrac{1}{1+2+3}\right)\times\left(1+\dfrac{1}{1+2+3+4}\right)\times...\times\left(1+\dfrac{1}{1+2+3+...+997}\right)\)

#Toán lớp 6

0

HP

Hồng Phong Đoàn

2 tháng 5 2023 - olm

1) Tính tổng C = \(\left(1-\dfrac{1}{2}\right)\).\(\left(1-\dfrac{1}{3}\right)\).\(\left(1-\dfrac{1}{4}\right)\).....\(\left(1-\dfrac{1}{2022}\right)\) 2) Cho tổng A = \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) +...+ \(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\). Chứng tỏ rằng A...

Đọc tiếp

#Toán lớp 6

1

LS

Lê Song Phương

2 tháng 5 2023

1) Ta có

\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)

\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)

\(C=\dfrac{1}{2022}\)

2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)

Đúng(2)

NL

Nguyen Le Tuong Vy

12 tháng 7 2023 - olm

Tính:

a/\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(\dfrac{-1}{6}\right):\left(-3\right)\)

b/\(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)

#Toán lớp 7

3

H9

HT.Phong (9A5)

CTVHS

12 tháng 7 2023

a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)

\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)

\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{119}{720}\)

b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)

\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)

\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)

\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)

\(B=1\)

Đúng(2)

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

12 tháng 7 2023

a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)

A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)

A = \(\dfrac{119}{720}\)

b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]

B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]

B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]

B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))

B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)

B = 1

Đúng(4)

NP

Nguyễn Phương Chi

12 tháng 9 2021

Tính

a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}\)

b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}\)

#Toán lớp 7

2

NH

Nguyễn Hoàng Minh

12 tháng 9 2021

\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)

Đúng(2)

I

ILoveMath

12 tháng 9 2021

a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)

b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)

Đúng(0)

DT

Duong Thi Nhuong

18 tháng 4 2017

Tính rồi so A và B :

\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)

\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)

#Toán lớp 8

1

NN

Ngáo Nu

19 tháng 4 2017

\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)

\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)

\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)

\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)

Vì (2) > (1) => B > A

Đúng(0)

TN

Thơ Nụ =))

27 tháng 1

Tính G=5E-2F, biết

\(E=\dfrac{7}{3}.\dfrac{37}{3^2}.\dfrac{1297}{3^4}...\dfrac{6^2+1}{3^2};F=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{3^2}\right)\left(1+\dfrac{1}{3^4}\right)...\left(1+\dfrac{1}{3^{32}}\right)\)

#Toán lớp 8

0

NA

Nguyễn Anh Thư

6 tháng 10 2017

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...

Đọc tiếp

#Toán lớp 7

3

LT

Lê Thị Ngọc Duyên

7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)