a, \(\frac{-5}{8}-x:3\frac{5}{6}+7\frac{3}{4}=-2\) b, \(\frac{3}{4}:2\frac{4}{9}-\left|-3x+2\frac{2}{3}\right|=\frac{3}{4}\) c...

a, \(\frac{-5}{8}-x:3\frac{5}{6}+7\frac{3}{4}=-2\) b,\(\frac...

BX

bui xuan dieu

19 tháng 3 2020

Tìm x, biết:

a) \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{-27}{64}\)

b) \(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{265}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+3\right)^2}=\frac{64}{27}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

#Hỏi cộng đồng OLM #Toán lớp 7

1

S

santa

19 tháng 3 2020

a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

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M

Mèo_Hanna

18 tháng 10 2018 - olm

Bài 1: Thực hiện phép tính:a, \(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)b, \(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)c, \(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}\)d, \(\frac{4^{2002}.9^{1001}}{16^{1001}.3^{2003}}\)e, \(\sqrt{25-16}-\left|-3,7+0,7\right|\)Bài 2: Tìm xa, \(\frac{1}{3}x+\frac{4}{5}=3\frac{4}{5}\)b, \(\left|x+\frac{3}{4}\right|-2,25=1\frac{3}{4}\)c, \(\left(-x+\frac{2}{5}\right)^4=\frac{1}{16}\)d,...

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9

KS

kudo shinichi

18 tháng 10 2018

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

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YM

Yuki Min JinWon

18 tháng 10 2018

1

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

= \(3\cdot25:\frac{5}{4}\)

= \(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) =

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TP

tran phuong anh

25 tháng 7 2017 - olm

\(3,2.\frac{15}{16}-\left(75\%+\frac{2}{7}\right):\left(-1\frac{1}{28}\right)\)\(\left(0,25+12,5-\frac{5}{16}\right):\left[12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right]\)\(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)\(14,5-\frac{8}{9}:\left(35-34\frac{8}{9}\right).\frac{9}{8}\)\(1\frac{1}{15}-\left(\frac{1}{15}+\frac{4}{9}:\frac{-2}{3}-\frac{28}{16}.\frac{6}{35}\right)-\frac{3}{10}\)Tìm...

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#Hỏi cộng đồng OLM #Toán lớp 7

11

HT

Hoàng Thảo

25 tháng 7 2017

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

\(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

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HT

Hoàng Thảo

25 tháng 7 2017

giờ mk phải đi ngủ r mai mk làm cho

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NT

Nguyễn Thanh Bình

1 tháng 8 2019

bài 2:tìm x giá trị biểu thức a)2-\(|^{ }_{ }\)3x-\(\frac{1}{4}|\)=\(\left|-\frac{5}{4}\right|\) f)-2,5:\(\left|\frac{3}{4}x+\frac{1}{2}\right|=3\) b)5-3.\(\left|5-2x\right|\)=-4,5 g)\(\left|2x-5\right|=\left|x=1\right|\) c)\(\frac{3}{2}+\frac{3}{4}.\left|x-\frac{3}{4}\right|=\frac{7}{4}\) d)\(\frac{11}{4}+\frac{3}{2}\left|4x-\frac{1}{5}\right|=\frac{7}{2}\) e)\(\frac{21}{5}+3:\left|\frac{x}{4}-\frac{2}{3}\right|=6\) ...

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#Hỏi cộng đồng OLM #Toán lớp 7

8

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

3 tháng 9

Câu g đề bài lỗi nhé e

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NT

Nguyễn Thị Thương Hoài

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3 tháng 9

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LN

Linh Nguyễn

26 tháng 10 2016 - olm

1.Tìm x biếta) \(\left(\frac{2}{5}-x\right)\):\(1\frac{1}{3}+\frac{1}{2}=-4\)b) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)c) \(\frac{-3x}{4}.\left(\frac{1}{x}+\frac{2}{7}\right)=0\)d) 3-\(\frac{1-\frac{1}{2}}{1+\frac{1}{x}}=2\frac{2}{3}\)e)\(\left(2x+\frac{1}{3}\right)^2=\frac{9}{16}\)f) \(4^x+4^{x+3}=4160\)g) \(\left(x-2,5\right):1\frac{1}{2}=x:2\)h) \(\left(\frac{1}{3}+x\right):\left(x+\frac{3}{4}\right)=\frac{7}{9}\)i)...

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#Hỏi cộng đồng OLM #Toán lớp 7

2

NT

nguyen thi thanh thao

26 tháng 10 2016

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)

\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

\(\frac{2}{5}-x=-3\)

\(x=\frac{2}{5}-\left(-3\right)\)

\(x=\frac{2}{5}+3\)

\(x=\frac{3}{5}-\frac{15}{5}\)

\(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\)

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NT

nguyen thi thanh thao

26 tháng 10 2016

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

\(-3+\frac{3}{x}=\frac{-25}{12}\)

\(\frac{3}{x}=\frac{-25}{12}+3\)

\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

\(\frac{3}{x}=\frac{5}{6}\)

\(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ

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VT

Vu Thanhh Dat

6 tháng 7 2019

...

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#Hỏi cộng đồng OLM #Toán lớp 7

1

NL

Ngọc Lan Tiên Tử

6 tháng 7 2019

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)

=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)

=> \(-\frac{3}{4}+\left(-2x\right)=-2\)

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

Vậy \(x\in\left\{\frac{5}{8}\right\}\)

\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)

=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)

=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)

=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)

=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)

Vậy \(x\in\left\{-\frac{39}{40}\right\}\)

\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)

=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)

( chiệt tiêu )

=> \(5x-6x+26=-14-7x\)

=> \(-x+26=-14-7x\)

=> \(-x+7x=-14-26\)

=> \(6x=-40\)

=> \(x=-40:6=\frac{20}{3}\)

Vậy \(x\in\left\{\frac{20}{3}\right\}\)

\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)

=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)

( chiệt tiêu )

=> \(2\left(2x-3\right)-9=5-3x-2\)

=> \(4x-6-9=3-3x\)

=> \(4x-15=3-3x\)

=> \(4x+3x=3+15\)

=> \(7x=18\)

=> \(x=18:7=\frac{18}{7}\)

Vậy \(x\in\left\{\frac{18}{7}\right\}\)

\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)

ĐKXĐ : \(x\ne0\)

=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)

=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)

=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)

=> \(\frac{32}{3x}=\frac{1}{4}\)

=> \(3x=32.4:1=128\)

=> \(x=128:3=\frac{128}{3}\)

Vậy \(x\in\left\{\frac{128}{3}\right\}\)

\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)

ĐKXĐ :\(x\ne1;\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)

=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)

=> \(\frac{26+5-2}{2\left(x-1\right)}\)

=> \(\frac{29}{2\left(x-1\right)}\)

\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)

=> \(x=\frac{19}{10}:2=\frac{19}{20}\)

Vậy \(x\in\left\{\frac{19}{20}\right\}\)

\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)

=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)

=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)

=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{1}{4}\right\}\)

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TK

Trần Khởi My

22 tháng 10 2018 - olm

BT1:...

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Nguyễn Thị Thương Hoài

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PT

pham thi hoa

27 tháng 9 2020

...

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NC

Nguyễn Châu Mỹ Linh

20 tháng 6 2018 - olm

...

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\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

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