`x:(-5/3) -(-2/5) =3/10`

`=> x: (-5/3) + 2/5 =3/10`

`=> x:(-5/3) = 3/10-2/5`

`=>x:(-5/3)= 3/10 - 4/10`

`=> x:(-5/3) = -1/10`

`=> x= -1/10 * (-5/3)`

`=> x= 5/30`

`=>x= 1/6`

Vậy `x=1/6`

Lời giải:
$A=\frac{1}{2}+\frac{3}{2}+(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2012}$

$\frac{3}{2}A=\frac{3}{4}+(\frac{3}{2})^2+(\frac{3}{2})^3+(\frac{3}{2})^4+...+(\frac{3}{2})^{2013}$

$\Rightarrow \frac{3}{2}A-A=(\frac{3}{2})^{2013}+\frac{3}{4}-\frac{1}{2}-\frac{3}{2}$

$\Rightarrow \frac{1}{2}A=(\frac{3}{2})^{2013}-\frac{5}{4}$

$A=2(\frac{3}{2})^{2013}-\frac{5}{2}$

$\Rightarrow B-A=(\frac{3}{2})^{2013}-2(\frac{3}{2})^{2013}+\frac{5}{2}=\frac{5}{2}-(\frac{3}{2})^{2013}$

\(\dfrac{3}{x-5}=\dfrac{-4}{x-2}\left(x\notin\left\{5;2\right\}\right)\)

\(\Rightarrow3\left(x-2\right)=-4\left(x-5\right)\)

\(\Rightarrow3x-6=-4x+20\)

\(\Rightarrow3x+4x=20+6\)

\(\Rightarrow7x=26\)

\(\Rightarrow x=\dfrac{26}{7}\) (thỏa) 

_________

\(\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-39}{91}\left(x\ne0\right)\)

\(\Rightarrow\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-3}{7}\)

+) \(\dfrac{3}{x}=-\dfrac{3}{7}\)

\(\Rightarrow x=\dfrac{3\cdot-7}{3}=-7\) (thỏa) 

+) \(\dfrac{y}{28}=-\dfrac{3}{7}\)

\(\Rightarrow y=\dfrac{28\cdot-3}{7}=-12\)

\(\left(-125\right)\cdot\left(-14\right)\cdot\left(-8\right)\cdot\left(-3\right)\)

\(=\left[\left(-125\right)\cdot\left(-8\right)\right]\cdot\left[\left(-14\right)\cdot\left(-3\right)\right]\)

\(=\left(125\cdot8\right)\cdot\left(14\cdot3\right)\)

\(=1000\cdot42\)

\(=42000\)

Ta có tính chất: \(\dfrac{a}{b}< \dfrac{a+m}{b+m}\) 

\(N=\dfrac{10^{20}+1}{10^{21}+1}< \dfrac{10^{20}+1+9}{10^{21}+1+9}\)

\(\Rightarrow N< \dfrac{10^{20}+10}{10^{21}+10}\)

\(\Rightarrow N< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\)

\(\Rightarrow N< \dfrac{10^{19}+1}{10^{20}+1}\) 

\(\Rightarrow N< M\)

Ta có tính chất: \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)

\(A=\dfrac{2022^{99}-1}{2022^{100}-1}>\dfrac{2022^{99}-1-2021}{2022^{100}-1-2021}\)

\(A>\dfrac{2022^{99}-2022}{2022^{100}-2022}\)

\(A>\dfrac{2022\left(2022^{98}-1\right)}{2022\left(2022^{99}-1\right)}\)

\(A>\dfrac{2022^{98}-1}{2022^{99}-1}\)

\(A>B\)