1) Tính $\dfrac{7^4.3-7^3}{7^4.6-7^3.2}$

NL

nguyễn lê bảo trâm

8 tháng 1 2021

1) Tính $\dfrac{7^4.3-7^3}{7^4.6-7^3.2}$ ; $\dfrac{10^3+5.10^2+5}{6^3+3.6^2+3^2}$ ; $E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}$ 2) Tìm x biết$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}$ ; $3^{x+1}+3^{x+3}=810$ MN ƠI ! GIÚP MIK VS > ....

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

8 tháng 1 2021

Bài 1:

a) Ta có: $\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}$

$=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}$

$=\dfrac{1}{2}$

c) Ta có: $E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}$

$\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}$

$\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)$

$\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}$

$\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}$

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NL

nguyễn lê bảo trâm

9 tháng 1 2021

thanks

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R

rinjin

8 tháng 10 2023

tính
a) A = $\dfrac{6^3+3.6^2+3^3}{-13}$
b) B =$\dfrac{2.8^4.27^2+4.6^9}{2^3.6^7+2^7.40.9^4}$
c) C=$\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{16}.6^{19}-7.2^{29}.27^6}$

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

8 tháng 10 2023

a: $A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27$

b: $B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}$

$=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}$

$=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}$

c: $C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}$

$=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}$

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TL

thuỳ linh

22 tháng 2 2023

tính giá trị
a,$\dfrac{\left(-3\right)^7.2^8}{6^7}$
b,$\dfrac{-3.7^4+7^3}{7^5.6-7^3.2}$
c,$\dfrac{5^3.3^5}{5^3.0.5+125.2.5}$

#Toán lớp 7

1

ED

🐟 ⋆ 🐇 🎀 𝒩𝑔𝓊𝓎ễ𝓃 Đă𝓃𝑔 𝒩𝒽â𝓃 🎀 🐇 ⋆ 🐟

22 tháng 2 2023

a) $\dfrac{\left(-3\right)^7\cdot2^8}{6^7}$

$=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2$

b) $\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}$

$=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}$

$=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}$

$=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}$

b) $\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}$

$=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}$

$=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}$

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NT

nguyen tu linh

24 tháng 6 2016

TINH NHANH :

A= ( -1/2)-(-3/5)+(-1/9)+1/131-(-2/7)+4/35-7/18

B=1/99-1/99.98-1/98.97-...-1/4.3-1/3.2-1/2.1

#Toán lớp 6

0

AS

★彡 A͛r͛a͛k͛i͛ S͛e͛i͛j͛u͛r͛o͛ 彡★

11 tháng 2 2018

Thực hiện phép tính:

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$B=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\dfrac{2^2}{7^2}-\dfrac{4}{343}}$

#Toán lớp 7

3

D

Despacito

11 tháng 2 2018

$B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}$

$B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}$

$B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}$

$B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}$

$B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}$

$B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}$

$B=\frac{300}{343}:\frac{1347}{343}$

$B=\frac{100}{449}$

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M

Mafia

11 tháng 2 2018

$A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}$

$A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}$

$A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}$

$A=\frac{-1}{2}+\frac{1710}{9}$

$A=\frac{-1}{2}+190$

$A=\frac{-1}{2}+\frac{380}{2}$

$A=\frac{379}{2}$

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AT

ánh tuyết nguyễn

29 tháng 1 2023

Tính các giới hạn sau:

a) $\lim\limits\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}$

b) $\lim\limits\dfrac{1}{\sqrt{3n+2}-\sqrt{2n+1}}$

c) $\lim\limits\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}$

#Toán lớp 11

3

AH

Akai Haruma

Giáo viên

29 tháng 1 2023

a.

$A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}$

Ta thấy:

$\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty $

$\lim (1+\frac{12}{n})=1$

Suy ra $A=\infty$

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AH

Akai Haruma

Giáo viên

29 tháng 1 2023

b.

$B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}$

$=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}$

Ta thấy:

$\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0$

$\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty$

$\Rightarrow B=\infty$

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:)))

30 tháng 3 2022

Tính nhanh

1/3.2/7+1/3.4/7+1/3:7

#Toán lớp 4

4

M

Mạnh=_=

30 tháng 3 2022

0.13427871148

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SM

Sơn Mai Thanh Hoàng

30 tháng 3 2022

$\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{1}{3}\times1=\dfrac{1}{3}$

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TN

Tú Nguyễn Văn

11 tháng 2 2018

Thực hiện phép tính:

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$B=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\dfrac{2^2}{7^2}-\dfrac{4}{343}}$

#Toán lớp 7

1

GH

Gia Hân Ngô

11 tháng 2 2018

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}$

$=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}$

= $\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}$

= $\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}$

= $\dfrac{7}{2}$

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GJ

GOT7 JACKSON

10 tháng 10 2018

1. Tính nhanh giá trị biểu thức: A = $\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}$ B = $\dfrac{1}{5}-\dfrac{2}{7}-\left(-\dfrac{2}{3}\right)+\dfrac{1}{42}-\dfrac{1}{4}-\dfrac{13}{28}+\dfrac{2}{15}$ C = $\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}$ D =...

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#Toán lớp 7

2

TH

Trịnh Hoài Thương

10 tháng 10 2018

A= $\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}$

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NL

Nguyễn Lê Phước Thịnh

12 tháng 10 2022

$B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)$

$=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}$

=1-82/84

=2/84=1/42

$C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)$

$=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)$

$=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}$

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

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TL

Trần Linh Chi

17 tháng 6 2017

a) $\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}$$-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}$ b)$\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}$ c)...

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#Toán lớp 7

4

DH

Đức Hiếu

17 tháng 6 2017

a,$\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}$

$=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}$

$=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}$

câu b và c xem lại đề nha

Chúc bạn học tốt!!!

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TL

Trần Linh Chi

17 tháng 6 2017

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