a) Tìm dạng phân số của x biết :(ghi cả cách ấn máy tính)\(x \frac{2006}{13 \frac{2}{23 \frac{5}{45 \frac{8}{9}}}}=\frac{2006}{45 \frac{2}{23 \frac{5}{13 \frac{8}{9}}}}\)

CB

con bạn thân

8 tháng 11 2018

dạy cho cj với

Tính hợp lý giá trị của biểu thức...

KL

Khánh Linh

8 tháng 5 2016

Đúng ghi Đ, sai ghi S vào chỗ chấm:a) \(\frac{3}{7}+\frac{1}{3}x\frac{3}{7}=\frac{16}{21}x\frac{3}{7}=\frac{16}{49}\) ......b)\(\frac{5}{9}+\frac{1}{5}x\frac{5}{6}=\frac{5}{9}x\frac{1}{6}=\frac{13}{18}\) ......c)\(\frac{13}{9}-\frac{7}{9}:\frac{2}{3}=\frac{6}{9}:\frac{3}{2}=1\) ......d)\(\frac{11}{8}-\frac{5}{8}:\frac{3}{4}=\frac{11}{8}-\frac{5}{8}x\frac{4}{3}=\frac{11}{8}-\frac{10}{24}=\frac{23}{24}\) ...

0

NA

Nguyễn Anh Thư

3 tháng 8 2017

#Toán lớp 4

6

MH

Mai Hồng Ngọc

3 tháng 8 2017

a Đ

b S

c S

d Đ

Đúng(1)

VT

Vũ Thị Thu Hằng

3 tháng 8 2017

a ) S

b ) Đ

c ) S

d ) Đ

k cho mk nhé

TD

Tứ diệp thảo mãi mãi yêu TFBOYS

21 tháng 1 2018

tìm x, biết:

\(\left(\frac{5}{12}+\frac{-5}{7}+\frac{-22}{45}+\frac{7}{12}+\frac{-23}{45}\right).\)I x I -9 = - 2

1

AO

Aoi Ogata

21 tháng 1 2018

\(\left(\frac{5}{12}-\frac{5}{7}-\frac{22}{45}+\frac{7}{12}-\frac{23}{45}\right).\left|x\right|-9=-2\)

\(\left(\frac{12}{12}-\frac{5}{7}-\frac{45}{45}\right).\left|x\right|=-2+9\)

\(\left(1-\frac{5}{7}-1\right).\left|x\right|=7\)

\(\frac{-5}{7}.\left|x\right|=7\)

\(\left|x\right|=7:\left(\frac{-5}{7}\right)\)

\(\left|x\right|=\frac{-49}{5}\)

\(\Rightarrow x\in\varnothing\) vì trị tuyệt đối của 1 số luôn dương

tính hợp lý giá trị các biểu thức sau :A= \(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)B= \(71\frac{38}{45}-\left(43\frac{8}{45}-1\frac{17}{57}\right)\)C= \(\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)D...

AJ

agelina jolie

6 tháng 6 2016

16

DQ

Đặng Quỳnh Ngân

6 tháng 6 2016

e) \(E=0,7.2\frac{2}{3}.20.0,375.\frac{5}{28}\)

\(=\left(0,7.20\right)\left(2\frac{2}{3}.0,375\right)\frac{5}{28}\)

\(=14.1.\frac{5}{28}\)

\(=14.\frac{5}{28}\)

\(=\frac{5}{2}\)

f) \(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)

\(=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)

\(=\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\frac{15}{78}\)

\(=\frac{39}{4}.40.\frac{15}{78}\)

\(=390.\frac{15}{78}\)

\(=78\)

Chúc bạn học tốt môn Toán!!!

PT

Phạm Tuấn Kiệt

6 tháng 6 2016

Mình ko ghi đề đâu

\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=29\frac{25}{32}\)

\(B=71\frac{38}{45}-43\frac{8}{45}+1\frac{17}{57}=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}=28\frac{2}{3}+1\frac{17}{57}=29\frac{55}{57}\)

\(C=-\frac{3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}=-\frac{3}{7}.1+2\frac{3}{7}=-\frac{3}{7}+2\frac{3}{7}=2\)

L

Luffy123

31 tháng 12 2018

Giải phương trình sau :

2) \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)

3) \(\frac{x+2}{13}+\frac{2x-45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)

#Toán lớp 8

12

S

shitbo

31 tháng 12 2018

\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)

\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)

\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)

\(.....................\)

đến đây thì dễ rồi :)

L

Luffy123

31 tháng 12 2018

mk không giải được phần sau

Tính nhanha) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\)b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}\)e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}\) Giúp mik vs mik đag cần...

TD

TÔ DIỆU LINH

21 tháng 7 2018

1

KT

Không Tên

21 tháng 7 2018

a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)

b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}-1+1=\frac{1}{3}\)

d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)

e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)

Tính hợp lý các biểu thức sau: (tức tính nhanh)\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right].0,15-\frac{1}{4}\)\(E=\frac{13}{30}+\frac{28}{45}.2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)Nhanh nhé, mình...

TP

Trần Phương Uyên

5 tháng 5 2019

1

NT

nguyen thi hien

15 tháng 8 2019

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

Tính nhanh ( nếu có thể): \(a,-3^2+\left\{-52:[\left(-3\right)^2\right\}\) \(d,(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}):(\frac{1}{6}-\frac{1}{8}-\frac{1}{24})\) \(b,2\frac{3}{7}+(\frac{2}{9}-1\frac{3}{7})-\frac{5}{3}:\frac{1}{9}\)...

DP

Đỗ Phi Phi

8 tháng 4 2018

1

NL

Nguyễn Lê Phước Thịnh

29 tháng 6 2022

a: \(=-9+\left\{-52:9\right\}=-9+\dfrac{-52}{9}=-\dfrac{133}{9}\)

b: \(=\dfrac{17}{7}+\left(\dfrac{-76}{63}\right):15\)

\(=\dfrac{17}{7}-\dfrac{76}{63}\cdot\dfrac{1}{15}=\dfrac{317}{135}\)

e: \(=\dfrac{-5}{13}\cdot\dfrac{7}{3}-\dfrac{2}{7}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)

\(=\dfrac{5}{13}\left(-\dfrac{7}{3}+\dfrac{1}{7}\right)-\dfrac{2}{7}\cdot\dfrac{8}{13}\)

\(=\dfrac{5}{13}\cdot\dfrac{-46}{21}-\dfrac{16}{91}=\dfrac{-278}{273}\)

TM

Trần Minh Thắng

22 tháng 8 2020

...

3

XO

Xyz OLM

22 tháng 8 2020

1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

=> x = - 1

b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)

=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)

=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

=> x + 2010 = 0

=> x = -2010

c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)

=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)

=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)

=> x = -1900

d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)

=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)

=> x = -2028

TD

Tạ Đức Hoàng Anh

22 tháng 8 2020

1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

+ TH₁: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

+ TH₂: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1\)

2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

\(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)

\(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)

\(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

+ TH₁: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)

+ TH₂: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)

\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)

mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2010\)

3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)

\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

\(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

+ TH₁: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)

+ TH₂: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)

\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)

mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1900\)

4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)

\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

\(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

+ TH₁: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)

+ TH₂: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)

\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)

mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=104\)

5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)

\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

\(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

+ TH₁: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)

+ TH₂: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)

mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2028\)

Chúc bn hok tốt nha