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1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)
=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)
=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)
=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)
=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)
=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)
=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)
=\(\dfrac{1}{2007}.\left(-1\right)+0\)
=\(\dfrac{-1}{2007}\)
2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)
=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)
=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)
=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)
=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)
bí rồi
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
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 \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\).
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
bai 1
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)
\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)
Ta luôn có :|x-2009|\(\ge\)0(1)
Mà :2009-|x-2009|=x nên 2009\(\ge\)x(2)
Vì (1)và(2) nên ta có x \(\in\){0;1;2;3;4;5;...;2009}
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
\(5^5.20^3-5^4.20^3=20^3\left(5^5-5^4\right)=20^3.5^4\left(5-1\right)=20^3.5^4.4\)
\(5^5.20^3-5^4.20^3+5^7.4^5=5^3.4^3.5^4.4+5^7.4^5=5^7.4^4+5^7.4^5=5^7.4^4\left(1+4\right)=5^8.4^4\)
\(\left(20+5\right)^3.4^5=25^3.4^5=\left(5^2\right)^3.4^5=5^6.4^5\)
Vậy kết quả của biểu thức là: \(\frac{5^8.4^4}{5^6.4^5}=\frac{25.5^6.4^4}{5^6.4^4.4}=\frac{25}{4}\)
\(\frac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3.4^5}=\frac{25000000-5000000+80000000}{15625.4^5}=\frac{100000000}{16000000}=\frac{25}{4}\)
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: 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 \(x\in\left\{22.51;-0.18\right\}\)
a: \(=\dfrac{2008}{2007}-2009\cdot2-\dfrac{2009}{2007}+2009\cdot2\)
=-1/2007
b: \(=\dfrac{5^5\cdot5^3\cdot2^6-5^4\cdot5^3\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^8\cdot2^6-5^7\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^7\cdot2^6\left(5-1+2^4\right)}{5^6\cdot2^{10}}=\dfrac{5}{16}\cdot\dfrac{20}{1}=\dfrac{100}{16}=\dfrac{25}{4}\)