a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

\(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\dfrac{9}{4}\)

\(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

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

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

\(\dfrac{3}{2}x=\dfrac{9}{6}+\dfrac{2}{6}\) hoặc \(\dfrac{3}{2}x=-\dfrac{9}{6}+\dfrac{2}{6}\)

\(\dfrac{3}{2}x=\dfrac{11}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{-7}{6}\)

\(x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}\) hoặc \(x=\dfrac{-7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}\)

\(x=\dfrac{11}{9}\) hoặc \(x=-\dfrac{7}{9}\)

Vậy...

Mình sửa lại ( từ dòng 4 trở xuống ):

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

\(\dfrac{3}{2}x=\dfrac{2}{6}-\dfrac{9}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{2}{6}+\dfrac{9}{6}\)

\(\dfrac{3}{2}x=-\dfrac{7}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{11}{6}\)

\(x=-\dfrac{7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}\) hoặc \(x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}\)

\(x=\dfrac{-7}{9}\) hoặc \(x=\dfrac{11}{9}\)

Vậy...

2×(x-1)²=4

2×(x-1)²=2²

2×(x-1)=2

x-1=2:2

x-1=1

x=1+1

x=2

 Vậy x = 2

\(\Leftrightarrow\left(x-1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{matrix}\right.\)

\(x^2-2mx+m^2-1=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=m^2-1\end{matrix}\right.\)

Ta có :

\(x_1^2+x_2^2=4\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4\)

\(\Leftrightarrow2m^2-2\left(m^2-1\right)-4=0\)

\(\Leftrightarrow2m^2-2m^2+2-4=0\)

\(\Leftrightarrow-2=0\left(VL\right)\)

Vậy không có giá trị m để thỏa mãn đề bài.

1/4- 1/6

= 3/12- 2/12

= 1/12

1/12

1/12

\(3^{-2}.2^x+2^x.3=\dfrac{7}{36}\)

\(=>2^x\left(\dfrac{1}{9}+3\right)=\dfrac{7}{36}\)

\(=>2^x.\dfrac{28}{9}=\dfrac{7}{36}\)

\(=>2^x=\dfrac{1}{16}\)

\(=>2^x=2^{-4}\)

\(=>x=-4\)

`4/3 + ( x + 3) xx 2 - 1/2 = 27/2`

`=>  ( x + 3) xx 2 - 1/2 = 27/2-4/3`

`=>  ( x + 3) xx 2 - 1/2 =73/6`

`=>  ( x + 3) xx 2 =73/6 +1/2`

`=>  ( x + 3) xx 2 =38/3`

`=>x+3=38/3 xx 1/2`

`=>x+3=19/3`

`=>x=19/3-3`

`=>x= 10/3`

\(3x-4x^2+6-8x>x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2>0\Leftrightarrow\left(5x-1\right)\left(x+2\right)>0\)

TH1 : \(\left\{{}\begin{matrix}5x-1>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x>-2\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{5}\)

TH2 : \(\left\{{}\begin{matrix}5x-1< 0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x< -2\end{matrix}\right.\Leftrightarrow x< -2\)