\(\dfrac{1}{3}+\dfrac{2}{3}x=\dfrac{2}{5}\)

\(\dfrac{2}{3}x=\dfrac{1}{15}\)

\(x=\dfrac{1}{10}\)

x=1/10

S = \(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{6\cdot7\cdot8}\)

S = \(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}\)

S = \(\dfrac{1}{1\cdot2}-\dfrac{1}{7\cdot8}\)

S = \(\dfrac{1}{2}-\dfrac{1}{56}\)

S = \(\dfrac{27}{56}\)

Vì \(\dfrac{13}{n-3}\in Z\)

=> 13 chia hết cho n - 3

=> \(n-3\inƯ\left(13\right)\)

=> \(n-3\in\left\{-13;-1;1;13\right\}\)

=> \(n\in\left\{-10;2;4;16\right\}\)

Ta có:

\(\left|x-9\right|\ge0\forall x\)

=> \(A=\left|x-9\right|+1\ge1\forall x\)

Dấu "=" xảy ra khi \(\left|x-9\right|=0\)

=> x = 9

Vậy MinA = 1 khi x = 9

c) \(\left|3x-1\right|+2\dfrac{3}{4}=3\dfrac{1}{16}\)

\(\left|3x-1\right|=3\dfrac{1}{16}-2\dfrac{3}{4}\)

\(\left|3x-1\right|=\left(3-2\right)+\left(\dfrac{1}{16}-\dfrac{3}{4}\right)\)

\(\left|3x-1\right|=1+\dfrac{-11}{16}\)

\(\left|3x-1\right|=\dfrac{5}{16}\)

=> \(3x-1\in\left\{\dfrac{5}{16};\dfrac{-5}{16}\right\}\)

TH1:

\(3x-1=\dfrac{5}{16}\)

\(3x=\dfrac{21}{16}\)

\(x=\dfrac{7}{16}\)

TH2:

\(3x-1=\dfrac{-5}{16}\)

\(3x=\dfrac{11}{16}\)

\(x=\dfrac{11}{48}\)

Vậy \(x\in\left\{\dfrac{11}{48};\dfrac{7}{16}\right\}\)

b) \(\left(x^2-9\right)\left(3-5x\right)=0\)

TH1:

\(x^2-9=0\)

\(x^2=9\)

\(x^2=3^2=\left(-3\right)^2\)

=>\(x\in\left\{3;-3\right\}\)

TH2:

\(3-5x=0\)

\(5x=3\)

\(x=\dfrac{3}{5}\)

Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)

a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)

\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)

\(\dfrac{19}{20}x=1\)

\(x=\dfrac{20}{19}\)

Vậy \(x=\dfrac{20}{19}\)

\(5\in N\) 

\(-2\in Z\) 

\(-2\in Q\)

\(-37\in Q\) 

\(N\subset Z\subset Q\)

\(=\dfrac{2011}{2012}\cdot\left(3-3-\dfrac{2011}{2012}\right)-\dfrac{1}{2012}\cdot\dfrac{2011}{2012}\)

\(=\dfrac{2011}{2012}\cdot\left(-\dfrac{2011}{2012}\right)-\dfrac{1}{2012}\cdot\dfrac{2011}{2012}\)

\(=\dfrac{2011}{2012}\cdot\left(\dfrac{-2011}{2012}-\dfrac{1}{2012}\right)\)

\(=\dfrac{2011}{2012}\cdot\left(-1\right)\)

\(=-\dfrac{2011}{2012}\)