\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)

Vậy `x=41/35`

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\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)

Vậy `x=-2`

a)

-4/7 - x = 3/5 - 2x

2x - x = 3/5 + 4/7

x = 41/35

Vậy x = 41/35

b)

3/7.x - 2/3.x = 10/21

x(3/7 - 2/3) = 10/21

x.(-5/21) = 10/21

x = 10/21 : (-5/21) = -2

Vậy x = -2

82x+1 – 8x = 3584 

=> 8x+1 = 3584 

8x+1 = 84 

x = 4-1 

x = 3 

Chúc bạn học tốt 

\(\dfrac{5}{2}\) - \(\dfrac{3}{2}\) \(\times\)(\(x\) - 2) = 2

     \(\dfrac{3}{2}\) \(\times\)(\(x\) - 2) = \(\dfrac{5}{2}\) - 2

      \(\dfrac{3}{2}\) \(\times\) (\(x\) - 2) = \(\dfrac{1}{2}\)

               \(x\)  - 2  = \(\dfrac{1}{2}\) : \(\dfrac{3}{2}\)

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

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

                  \(x\)         = \(\dfrac{7}{3}\)

\(\dfrac{5}{2}-\dfrac{3}{2}\cdot\left(x-2\right)=2\\ \Rightarrow\dfrac{5}{2}-\dfrac{3}{2}x+\dfrac{6}{2}=2\\ \Rightarrow\dfrac{5}{2}-\dfrac{3}{2}x=2-3\\ \Rightarrow\dfrac{5}{2}-\dfrac{3}{2}x=-1\\ \Rightarrow\dfrac{3}{2}x=\dfrac{5}{2}-\left(-1\right)\\ \Rightarrow\dfrac{3}{2}x=\dfrac{7}{2}\\ \Rightarrow x=\dfrac{7}{2}:\dfrac{3}{2}\\ \Rightarrow x=\dfrac{7}{3}\)

(x - 13 + y)² + (x - 6 - y)² ≥ 0 + 0 = 0

Vì dấu "=" xảy ra nên x - 13 + y = 0 và x - 6 - y = 0

x + y = 13 và x - y = 6

x = (13 - 6) : 2 = 3,5

y = 13 - 3,5 = 9,5

Vậy x = 3,5 và y = 9,5

(\(x\) - 13 + y)² + (\(x\) - 6 - y)² = 0

(\(x\) - 13 + y)² ≥ 0 ∀ \(x;y\)

(\(x-6-y\))² ≥ 0 ∀ \(x;y\)

⇒(\(x-13+y\))² + (\(x\) - 6- y)² = 0

⇔ \(\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-6-y=0\\x-13+y+x-6-y=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}y=x-6\\2x=19\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{19}{2}-6\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)

|5 - \(\dfrac{2}{3}\)\(x\)| + |\(\dfrac{2}{3}\)y - 4| =0

|5 - \(\dfrac{2}{3}\)\(x\)| ≥ 0 ∀ \(x\); |\(\dfrac{2}{3}\)y - 4| ≥ 0 ∀ y

⇒ |5 - \(\dfrac{2}{3}\)\(x\)| + |\(\dfrac{2}{3}\)y - 4| = 0 ⇔ \(\left\{{}\begin{matrix}5-\dfrac{2}{3}x=0\\\dfrac{2}{3}y-4=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{15}{2}\\y=6\end{matrix}\right.\)

Chọn B

-1\(\dfrac{1}{2}\)  = - (1\(\dfrac{1}{2}\)) = -  \(\dfrac{2+1}{2}\) = - \(\dfrac{3}{2}\)

Chọn b, - \(\dfrac{3}{2}\)

\(\dfrac{5}{6}\) - (\(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) - \(x\)) = 10 - | \(\dfrac{1}{3}\) - \(\dfrac{1}{2}\)|

\(\dfrac{5}{6}\) - (\(\dfrac{13}{8}\)- \(x\)) = 10 - |-\(\dfrac{1}{6}\)|

 \(\dfrac{5}{6}\) - \(\dfrac{13}{8}\) + \(x\) = 10 + \(\dfrac{1}{6}\)

- \(\dfrac{19}{24}\) + \(x\)      = \(\dfrac{61}{6}\)

            \(x\)     = \(\dfrac{61}{6}\) + \(\dfrac{19}{24}\)

            \(x\)     = \(\dfrac{263}{24}\)