\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

\(a,\dfrac{x}{8}=\dfrac{7}{-2}\\ \Rightarrow x=-28\\ b,\dfrac{1-2x}{6}=\dfrac{-1}{2}\\ \Leftrightarrow2-4x=-6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\\ c,\dfrac{x+2}{3}=\dfrac{x+3}{4}\\ \Leftrightarrow4x+8=3x+9\\ \Leftrightarrow x=1\\ d,\dfrac{10}{2-x}=2\\ \Leftrightarrow4-2x=10\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

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

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

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

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

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)

b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)

⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)

⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)

⇒ \(x=13\)

c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)

⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)

⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)

⇒ \(x=10\)

d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

⇒ \(x=13\)

câu d chưa có đóng ngoặc kìa bn

Hoàng Tử Hà

Y

Dương Bá Gia Bảo

Nguyễn Thị Diễm Quỳnh

lê thị hương giang

Vũ Huy Hoàng

Nguyễn Thị Linh Chi

Luân Đào

DƯƠNG PHAN KHÁNH DƯƠNG

Nguyễn Ngân Hòa

Nguyễn Huy Tú

Lightning Farron

m.n làm giúp mình với ak

\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)

\(\dfrac{1}{2}.x+\dfrac{1}{2}=\dfrac{5}{2}\Leftrightarrow\dfrac{1}{2}.x=\dfrac{5}{2}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}.x=2\Leftrightarrow x=4\)

\(\dfrac{1}{2}-\dfrac{2}{3}.x=\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{1}{2}+\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{13}{12}\Leftrightarrow x=\dfrac{13}{12}:\dfrac{2}{3}\Leftrightarrow x=\dfrac{13}{8}\)