\(A=\dfrac{1}{2}+\left|2x-1\right|\)

\(B=...">

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25 tháng 10 2021

\(A=\dfrac{1}{2}+\left|2x-1\right|\ge\dfrac{1}{2}\forall x\)

\(minA=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{\left|x\right|+2007}{2008}\ge\dfrac{0+2007}{2008}=\dfrac{2007}{2008}\)

\(minB=\dfrac{2007}{2008}\Leftrightarrow x=0\)

2 tháng 10 2017

bai 1

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)

\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)

\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)

a: \(\Leftrightarrow-\dfrac{3}{2x-3}=\dfrac{2}{5}-\dfrac{3}{2}-3=\dfrac{-41}{10}\)

=>41(2x-3)=30

=>82x-123=30

=>82x=153

hay x=153/82

b: \(\Leftrightarrow\left(x-1\right)\left(7-2x\right)=0\)

=>x=1 hoặc x=7/2

c: \(\Leftrightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+10}{2009}+1\right)+\left(\dfrac{x+11}{2008}+1\right)+\left(\dfrac{x+12}{2007}+1\right)\)

=>x+2019=0

hay x=-2019

a: \(=\dfrac{2008}{2007}-2009\cdot2-\dfrac{2009}{2007}+2009\cdot2\)

=-1/2007

b: \(=\dfrac{5^5\cdot5^3\cdot2^6-5^4\cdot5^3\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)

\(=\dfrac{5^8\cdot2^6-5^7\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)

\(=\dfrac{5^7\cdot2^6\left(5-1+2^4\right)}{5^6\cdot2^{10}}=\dfrac{5}{16}\cdot\dfrac{20}{1}=\dfrac{100}{16}=\dfrac{25}{4}\)

a: TH1: x>=0

=>x+x=1/3

=>x=1/6(nhận)

TH2: x<0

Pt sẽ là -x+x=1/3

=>0=1/3(loại)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)

c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)

\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)

\(\Leftrightarrow3x^2-63x+60=4x+72\)

=>3x^2-67x-12=0

hay \(x\in\left\{22.51;-0.18\right\}\)

8 tháng 4 2017

1. a) \(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy \(x\le2009.\)

b) Ta có: \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y-z\right|\ge0\forall x,y,z\end{matrix}\right.\) \(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)

Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}=0\\\left(y-\dfrac{2}{5}\right)^{2008}=0\\\left|x+y-z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\).

8 tháng 4 2017

Bạn kia làm câu 1 rồi thì mình làm câu 2 nhé!

2. Ta có:\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)

\(\Rightarrow\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{5b-3c}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{15a-10b+6c-15a}{25+9}\)=\(\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Rightarrow\dfrac{-5b+3c}{17}=\dfrac{5b-3c}{2}\Rightarrow5b-3c=0\)

=> 5b=3c =>\(\left\{{}\begin{matrix}b=\dfrac{3}{5}c\\a=\dfrac{2}{5}c\end{matrix}\right.\)

=>\(\dfrac{3}{5}c+\dfrac{2}{5}c+c=-50\)

=> \(c\left(\dfrac{3}{5}+\dfrac{2}{5}+1\right)=-50\)

=> 2c = -50

=> c= -25

=>\(\left\{{}\begin{matrix}b=-25.\dfrac{3}{5}=-15\\a=-25.\dfrac{2}{5}=-10\end{matrix}\right.\)

Vậy a= -10; b= -15; c= -25

1 tháng 7 2017

a) ( x + 5 )3 = -64

x + 5 = - 4

x = - 4 - 5

x = -9

b) (2x - 3)2=9

2x - 3 = 3

2x = 3+3

2x = 6

x = 6 : 2

x = 3

e) \(\dfrac{8}{2x}=4\)

=> 4 . 2x = 8

8x =8

x = 8 : 8

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)

=> x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(\dfrac{1}{4}.x=\dfrac{1}{32}\)

x = \(\dfrac{1}{32}:\dfrac{1}{4}\)

x = \(\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\dfrac{-1}{27}\)

2 tháng 7 2017

a) (x + 5)3 = -64

=> (x + 5)3 = (-4)3

x + 5 = -4

x = -4 - 5

x = -9

b) (2x - 3)2 = 9

=> (2x - 3)2 = (\(\pm\)3)2

=> 2x - 3 = 3 hoặc 2x - 3 = -3

*2x - 3 = 3

2x = 3 + 3

2x = 9

x = \(\dfrac{9}{2}\)

*2x - 3 = -3

2x = -3 + 3

2x = 0

x = 0 : 2

x = 0

Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)

c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)

\(\dfrac{x}{2}=8\)

x = 8 : 2

x = 4

d) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

=> (-2)n . (-2)2= (-2)5

(-2)n = (-2)5 : (-2)2

(-2)n = (-2)3

Vậy n = 3

e) \(\dfrac{8}{2x}=4\)

=> 2x . 4 = 8

2x = 8 : 4

2x = 2

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)

2x - 1 = 3

2x = 3 + 1

2x = 4

x = 4 : 2

x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)

\(x=\left(\dfrac{1}{2}\right)^3\)

\(x=\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^3\)

\(x=\dfrac{-1}{27}\).

a: \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

nên \(\left\{{}\begin{matrix}2x-1=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=x+y=\dfrac{9}{10}\end{matrix}\right.\)

b: Bạn xem lại đề, nghiệm rất xấu

 

 

20 tháng 10 2017

Ta luôn có :|x-2009|\(\ge\)0(1)

Mà :2009-|x-2009|=x nên 2009\(\ge\)x(2)

(1)(2) nên ta có x \(\in\){0;1;2;3;4;5;...;2009}

20 tháng 10 2017

a)

\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy \(x\le2009\)

b)

Vì \(\left(2x+1\right)^{2008}\ge0\forall x\)

\(\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\)

\(\left|x+y-z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)

Mà theo đề bài :

\(\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

\(\Rightarrow\left(2x+1\right)^{2008}=0;\left(y-\dfrac{2}{5}\right)^{2008}=0;\left|x+y-z\right|=0\)

*) Với \(\left(2x+1\right)^{2008}=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\)

*) Với \(\left(y-\dfrac{2}{5}\right)^{2008}=0\)

\(\Rightarrow y-\dfrac{2}{5}=0\)

\(\Rightarrow y=\dfrac{2}{5}\)

*) Với \(\left|x+y-z\right|=0\)

\(\Rightarrow x+y-z=0\)

\(\Rightarrow\dfrac{-1}{2}+\dfrac{2}{5}-z=0\)

\(\Rightarrow\dfrac{-1}{10}-z=0\)

\(\Rightarrow z=\dfrac{-1}{10}\)

Vậy \(x=\dfrac{-1}{2};y=\dfrac{2}{5};z=\dfrac{-1}{10}\)

20 tháng 10 2017

a, 2009 - \(\left|x-2009\right|\) = x

=> \(\left|x-2009\right|\) = 2009 - x

=> \(\left[{}\begin{matrix}x-2009=2009-x\\x-2009=-2009-x\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4018\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2009\\x=0\end{matrix}\right.\)

Vậy x \(\in\)n { 2009 ; 0 }

4 tháng 11 2018

1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)

=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)

=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)

=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)

=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)

=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)

=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)

=\(\dfrac{1}{2007}.\left(-1\right)+0\)

=\(\dfrac{-1}{2007}\)

4 tháng 11 2018

2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)

=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)

=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)

=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)

=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)

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