\(\left\{{}\begin{matrix}5x=5m\\y=2x-m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=10-m+1=11-m\end{matrix}\right.\)

Thay vào ta đc 

\(2m^2-3\left(11-m\right)=2\Leftrightarrow2m^2-33+3m=2\Leftrightarrow2m^2+3m-35=0\Leftrightarrow m=\dfrac{7}{2};m=-5\)

a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)

hay \(x=\dfrac{1}{6}\)

Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)

b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)

\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)

Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)

\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

Vậy: \(B_{min}=4\) khi x=2 và y=6

a)\(x-15\%x=\frac{1}{3}\)

\(x.\left(1-15\%\right)=\frac{1}{3}\)

\(x.\frac{-280}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{-280}{3}\)

\(x=\frac{-1}{280}\)

Vậy \(x=\frac{-1}{280}\)

b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)

\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)

\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)

\(-\frac{17}{10}x=\frac{-61}{30}\)

\(x=\frac{-61}{30}:\frac{-17}{10}\)

\(x=\frac{61}{51}\)

Vậy \(x=\frac{61}{51}\)

x² + 2x = 0

=> x(x + 2) = 0

=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

(x - 2) + 3.x² - 6x = 0

=> (x - 2) + 3x² - 3x . 2 = 0

=> (x - 2) + 3x.(x - 2) = 0

=> (1 + 3x)(x - 2) = 0

=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)

so 9 la dấu ( đấy mk nham

so 9 la dau (

Bước 1: Tìm \(\Delta\)và rút gọn

Bước 2: Để pt .. <=> \(\Delta\).. 0

Bước 3: Kết luận

Chúc bạn thành công =))))))

Bổ sung thêm bước 2: Là phải giải bất pt hoặc pt nhé 

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{10}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

\(x=\frac{185}{4}\)

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)

\(\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

       \(x=\frac{37}{8}:\frac{1}{10}\)

      \(x=\frac{185}{4}\)

10 + (2x - 1) ² : 3 = 13

=>    (2x - 1) ² : 3 = 13 - 10

=>    (2x - 1) ² : 3 = 3

=>    (2x - 1) ²      =  3 . 3 

=>    (2x - 1) ²      =  3 ²  

=>              2x - 1 = 3 

=>                   2x = 3 + 1 

=>                   2x = 4

=>                      x = 2

10 + (2x - 1)² : 3 = 13 

=> (2x - 1)² : 3 = 13 - 10

=> (2x - 1 )² : 3 = 3

=>  (2x - 1)²      = 9

=>  (2x - 1)²      = 3²

=>  2x  - 1         = 3

 => 2x                = 4

 => x    = 2

Vậy x = 2

\(10+\left(2x-1\right)^2:3=13\)

\(\left(2x-1\right)^2:3=13-10\)

\(\left(2x-1\right)^2:3=3\)

\(\left(2x-1\right)^2=3.3\)

\(\left(2x-1\right)^2=9\)

\(\left(2x-1\right)^2=\left(\pm3\right)^2\)

\(2x-1=\pm3\)

\(TH1:2x-1=3\)                                                                    \(TH2:2x-1=-3\)

\(2x=3+1\)                                                                                      \(2x=-3+1\)

\(2x=4\)                                                                                                \(2x=-2\)

\(x=4:2\)                                                                                                \(x=-2:2\)

\(x=2\)                                                                                                    \(x=-1\)

                                  Vậy \(x\in\left\{2,-1\right\}\)