$A=\dfrac{\left(\dfrac15-\dfrac27\right)\cdot\dfrac34-\dfrac34\cdot\left(\dfrac13-\dfrac27\right)}{\dfrac15\cdot\dfrac27-\dfrac13\cdot\left(\dfrac27+\dfrac39\right)+\dfrac39+\dfrac15}$

$=\dfrac{\dfrac34\left[\left(\dfrac15-\dfrac27\right)-\left(\dfrac13-\dfrac27\right)\right]}{\dfrac2{35}-\dfrac13\left(\dfrac27+\dfrac13\right)+\dfrac13+\dfrac15}$

$=\dfrac{\dfrac34\left(\dfrac15-\dfrac13\right)}{\dfrac2{35}-\dfrac13\cdot\dfrac{13}{21}+\dfrac13+\dfrac15}$

$=\dfrac{\dfrac34\cdot\left(-\dfrac2{15}\right)}{\dfrac2{35}-\dfrac{13}{63}+\dfrac{8}{15}}$

$=\dfrac{-\dfrac1{10}}{\dfrac{18-65+168}{315}}$

$=\dfrac{-\dfrac1{10}}{\dfrac{121}{315}}$

$=-\dfrac1{10}\cdot\dfrac{315}{121}$

$=-\dfrac{63}{242}.$

( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0

 <=> +) 1/7 . x - 2/7 = 0                                    +)    (- 1 / 5) . x +3/5 = 0                              +)  1/ 3 . x + 4/ 3 = 0

                    x = 2                                                                  x = 3                                                         x = 4

                                                    Vậy x = 2 : x = 3 ; x=4

$\left(\dfrac17x-\dfrac27\right)\left(-\dfrac15x+\dfrac35\right)\left(\dfrac13x+\dfrac43\right)=0$

$\Leftrightarrow\dfrac{x-2}{7}\cdot\dfrac{3-x}{5}\cdot\dfrac{x+4}{3}=0$

$\Leftrightarrow x-2=0\ \text{hoặc}\ 3-x=0\ \text{hoặc}\ x+4=0$

$\Leftrightarrow x=2\ \text{hoặc}\ x=3\ \text{hoặc}\ x=-4.$

Vậy $\boxed{x\in\{-4,\ 2,\ 3\}.}$

a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)

b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10

a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)

b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)

c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)

d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)

e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)

\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)

$\textbf{a)}$

$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$

$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$

$=-\dfrac{13}{8}+\dfrac{39}{56}$

$=-\dfrac{13}{14}.$

$\textbf{b)}$

$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$

$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$

$=\dfrac19.$

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