\(a.\dfrac{1}{2}-3x=-\dfrac{2}{5}\\ 3x=\dfrac{1}{2}+\dfrac{2}{5}\\ 3x=\dfrac{9}{10}\\ x=\dfrac{9}{10}:3\\ x=\dfrac{3}{10}\\ b.-x+\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{1}{2}+\dfrac{5}{6}\\ x=\dfrac{4}{3}\\ c.x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\\ x+\dfrac{3}{5}=\dfrac{4}{25}\\ x=\dfrac{4}{25}-\dfrac{3}{5}\\ x=-\dfrac{11}{25}\\ d.\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\\ x=\dfrac{1}{7}:-\dfrac{3}{14}=-\dfrac{2}{3}\\ e.-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\\ \dfrac{1}{7}-x=\dfrac{1}{21}:-\dfrac{1}{3}=-\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\\ h.\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\\ \dfrac{1}{4}-3x+\dfrac{3}{2}=-\dfrac{3}{4}\\ 3x=\dfrac{1}{4}+\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{5}{2}\\ x=\dfrac{5}{2}:3\\ x=\dfrac{5}{6}\\ i.\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\\ \dfrac{2}{3}+2x=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\\ 2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{23}{21}\\ x=\dfrac{-23}{21}:2=-\dfrac{23}{42}\)

a: \(\dfrac{1}{2}-3x=-\dfrac{2}{5}\)

=>\(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{5}{10}+\dfrac{4}{10}=\dfrac{9}{10}\)

=>\(x=\dfrac{9}{10}:3=\dfrac{9}{30}=\dfrac{3}{10}\)

b: \(-x+\dfrac{1}{2}=-\dfrac{5}{6}\)

=>\(-x=-\dfrac{5}{6}-\dfrac{1}{2}=-\dfrac{5}{6}-\dfrac{3}{6}=-\dfrac{8}{6}=-\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}\)

c: \(x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\)

=>\(x+\dfrac{3}{5}=\dfrac{4}{25}\)

=>\(x=\dfrac{4}{25}-\dfrac{3}{5}=\dfrac{4}{25}-\dfrac{15}{25}=-\dfrac{11}{25}\)

d: \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)

=>\(\dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\)

=>\(x=-\dfrac{1}{7}:\dfrac{3}{14}=-\dfrac{1}{7}\cdot\dfrac{14}{3}=-\dfrac{2}{3}\)

e: \(-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\)

=>\(\dfrac{1}{7}-x=\dfrac{1}{21}:\dfrac{-1}{3}=\dfrac{-1}{21}\cdot3=-\dfrac{1}{7}\)

=>\(x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\)

h: \(\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\)

=>\(-3x+\dfrac{5}{4}=-\dfrac{3}{4}\)

=>\(-3x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)

=>\(x=\dfrac{-2}{-3}=\dfrac{2}{3}\)

i: \(\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\)

=>\(2x+\dfrac{2}{3}=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\)

=>\(2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{9}{21}-\dfrac{14}{21}=-\dfrac{23}{21}\)

=>\(x=-\dfrac{23}{21}:2=-\dfrac{23}{42}\)

\(a.\dfrac{3}{7}\cdot\dfrac{5}{8}+\dfrac{3}{7}\cdot\dfrac{11}{8}+\dfrac{11}{7}=\dfrac{3}{7}\cdot\left(\dfrac{5}{8}+\dfrac{11}{8}\right)+\dfrac{11}{7}=\dfrac{3}{7}\cdot2+\dfrac{11}{7}=\dfrac{6}{7}+\dfrac{11}{7}=\dfrac{17}{7}\\ b.\dfrac{3}{8}\cdot19\dfrac{1}{3}-\dfrac{3}{8}\cdot\left(33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot-14=\dfrac{-21}{8}\\ c.\dfrac{1}{3}\cdot\dfrac{5}{4}+\dfrac{1}{3}\cdot\dfrac{7}{4}-2022^0=\dfrac{1}{3}\cdot\left(\dfrac{5}{4}+\dfrac{7}{4}\right)-1=\dfrac{1}{3}\cdot\dfrac{12}{4}-1=\dfrac{1}{3}\cdot3-1=1-1=0\\ d.\dfrac{5}{13}+\left(-\dfrac{5}{17}\right)+\dfrac{-21}{41}+\dfrac{8}{13}+\dfrac{-20}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(-\dfrac{5}{17}\right)+\left(\dfrac{-21}{41}+\dfrac{-20}{41}\right)=1+\left(-\dfrac{5}{17}\right)-1=-\dfrac{5}{17}\)

\(e.\dfrac{27}{13}:\dfrac{9}{7}+\dfrac{12}{13}:\dfrac{9}{7}=\dfrac{27}{13}\cdot\dfrac{7}{9}+\dfrac{12}{13}\cdot\dfrac{7}{9}=\dfrac{7}{9}\cdot\left(\dfrac{27}{13}+\dfrac{12}{13}\right)=\dfrac{7}{9}\cdot\dfrac{39}{13}=\dfrac{7}{9}\cdot3=\dfrac{7}{3}\\ g.\dfrac{8}{15}\cdot-\dfrac{4}{9}+\dfrac{8}{15}:\dfrac{-9}{5}-3\dfrac{2}{5}=\dfrac{8}{15}\cdot\dfrac{-4}{9}+\dfrac{8}{15}\cdot\dfrac{-5}{9}-\dfrac{17}{5}=\dfrac{8}{15}\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)-\dfrac{17}{5}=-\dfrac{8}{15}-\dfrac{17}{5}=-\dfrac{59}{15}\\ h.\left(-\dfrac{2}{3}+\dfrac{3}{13}\right):\dfrac{7}{8}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right):\dfrac{7}{8}=\left(-\dfrac{2}{3}+\dfrac{3}{13}\right)\cdot\dfrac{8}{7}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right)\cdot\dfrac{8}{7}=\dfrac{8}{7}\cdot\left(-\dfrac{2}{3}+\dfrac{3}{13}-\dfrac{1}{3}+\dfrac{10}{3}\right)=\dfrac{8}{7}\cdot\left(-1+1\right)=\dfrac{8}{7}\cdot0=0\)

\(a.\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:-\dfrac{9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\cdot\dfrac{-2}{9}=\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

\(b.\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{4\cdot2-5}{10}=\dfrac{1}{2}:\dfrac{3}{10}=\dfrac{1}{2}\cdot\dfrac{10}{3}=\dfrac{5}{3}\)

\(c.\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}:\dfrac{5}{9}-1=\dfrac{1}{9}\cdot\dfrac{9}{5}-1=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

\(d.\left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}-6\dfrac{1}{2}\right)=\dfrac{9}{25}-\dfrac{4}{5}+6\dfrac{1}{2}=\dfrac{9}{25}-\dfrac{20}{25}+\dfrac{13}{2}=\dfrac{-11}{25}+\dfrac{13}{2}=\dfrac{303}{50}\)

a) \(\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:\dfrac{-9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\times\dfrac{2}{-9}\\ =\dfrac{1}{4}-\left(-\dfrac{1}{12}\right)=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

b) \(\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{3}{10}\\ =\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

c) \(\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}\times\dfrac{9}{5}-1=\dfrac{1}{5}-1\\ =-\dfrac{4}{5}\)

\(\left(2\dfrac{3}{4}-\dfrac{1}{2}\right):\dfrac{3}{5}-3\)

\(=\left(2+\dfrac{1}{4}\right):\dfrac{3}{5}-3=\dfrac{9}{4}\cdot\dfrac{5}{3}-3\)

\(=\dfrac{45}{12}-3=\dfrac{45}{12}-\dfrac{36}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(12\cdot53+53\cdot47-53\cdot41\)

\(=53\cdot\left(12+47-41\right)\)

\(=53\cdot18=954\)

$12\times53+53\times47-53\times41$

$=53\times(12+47-41)$

$=53\times(59-41)$

$=53\times18=954$

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

b: A nằm giữa O và B

=>OA+AB=OB

=>AB+3=6

=>AB=3(cm)

c: Vì A nằm giữa O và B

nên AO và AB là hai tia đối nhau

=>AO và Ax là hai tia đối nhau

Trên tia AO, ta có: AO<AM

nên O nằm giữa A và M

=>AO+OM=AM

=>OM+3=6

=>OM=3(cm)

=>OM=OA(=3cm)

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

b: A nằm giữa O và B

=>OA+AB=OB

=>AB+3=6

=>AB=3(cm)

c: Vì A nằm giữa O và B

nên AO và AB là hai tia đối nhau

=>AO và Ax là hai tia đối nhau

Trên tia AO, ta có: AO<AM

nên O nằm giữa A và M

=>AO+OM=AM

=>OM+3=6

=>OM=3(cm)

=>OM=OA(=3cm)

Số lần khối lượng mặt trời gấp khối lượng trái đất là:

\(\dfrac{1998550\cdot10^{21}}{6\cdot10^{21}}=\dfrac{999275}{3}\left(lần\right)\)

Gọi x là thương của phép chia \(\left(x\inℕ\right)\)

Theo đề bai ta có:

\(a:28=x\) (dư 14)

\(\Rightarrow a=28\cdot x+14\)

\(\Rightarrow a=14\cdot\left(2\cdot x+1\right)\)

Nhận xét:

+) Vì \(14⋮2\) nên \(14\cdot\left(2\cdot x+1\right)⋮2\)

hay \(a⋮2\)

+) Vì \(14⋮14\) nên \(14\cdot\left(2\cdot x+1\right)⋮14\)

hay \(a⋮14\)

Vậy...

TH1: \(-5\le x\le2\)

=> \(\left(2-x\right)-4\left(5+x\right)=-23\)

\(=>2-x-20-4x=-23\)

\(=>-5x-18=-23\\ =>-5x=-23+18\\ =>-5x=-5\\ =>x=\dfrac{-5}{-5}=1\left(tm\right)\)

TH2: \(x>2\) 

\(\left(x-2\right)-4\left(5+x\right)=-23\) 

\(=>x-2-20-4x=-23\\ =>-3x-22=-23\\ =>-3x=-1\\ =>x=-\dfrac{1}{-3}=\dfrac{1}{3}\left(ktm\right)\)

TH3: \(x< -5\) 

\(\left(2-x\right)+4\left(5+x\right)=-23\\ =>2-x+20+4x=-23\\ =>3x+22=-23\\ =>3x=-45\\ =>x=-15\left(tm\right)\) 

Vậy: ...

$|2-x|-4|5+x|=-23$ (1)

+, Với $x< -5\Rightarrow \begin{cases} |2-x|=2-x\\|5+x|=-5-x \end{cases}$, (1) trở thành:

$2-x-4(-5-x)=-23$

$\Rightarrow 2-x+20+4x=-23$

$\Rightarrow 3x+22=-23$

$\Rightarrow 3x=-23-22$

$\Rightarrow 3x=--45$

$\Rightarrow x=-15$ (tmdk)

+, Với $-5\le x\le 2\Rightarrow \begin{cases} |2-x|=2-x\\|5+x|=5+x \end{cases}$, (1) trở thành:

$2-x-4(5+x)=-23$

$\Rightarrow 2-x-20-4x=-23$

$\Rightarrow -5x-18=-23$

$\Rightarrow -5x=-23+18$

$\Rightarrow -5x==-5$

$\Rightarrow x=1$ (tmdk)

+, Với $x>2\Rightarrow \begin{cases} |2-x|=x-2\\|5+x|=5+x \end{cases}$, (1) trở thành:

$x-2-4(5+x)=-23$

$\Rightarrow x-2-20-4x=-23$

$\Rightarrow -3x-22=-23$

$\Rightarrow -3x=-23+22$

$\Rightarrow -3x=-1$

$\Rightarrow  x=\frac13$ (loại)

Vậy: ...