GIÚP TỚ ĐI MẤY CẬU TỚ KO BT LÀM

Lời giải:
Xét tam giác $ABM$ và $ACM$ có:

$AB=AC$

$\widehat{BAM}=\widehat{CAM}$ (do $AM$ là tia phân giác $\widehat{A}$)

$AM$ chung

$\Rightarrow \triangle ABM=\triangle ACM$ (c.g.c)

$\Rightarrow BM=CM$

Hình vẽ

k = \(x.y\) = -2 x 30 = - 60

Chọn -60 

|3\(x\) - 1| +|1 - 3\(x\)| = 9

vì |3\(x\) - 1| = |1 - 3\(x\)| nên:

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

⇒2.|3\(x\) - 1| = 9

      |3\(x\) - 1| = \(\dfrac{9}{2}\)

      \(\left[{}\begin{matrix}3x-1=\dfrac{-9}{2}\\3x-1=\dfrac{9}{2}\end{matrix}\right.\)

       \(\left[{}\begin{matrix}3x=-\dfrac{9}{2}+1\\3x=\dfrac{9}{2}+1\end{matrix}\right.\)

        \(\left[{}\begin{matrix}3x=-\dfrac{7}{2}\\3x=\dfrac{11}{2}\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=\dfrac{11}{6}\end{matrix}\right.\)

       Vậy \(x\) \(\in\) {- \(\dfrac{7}{6}\); \(\dfrac{11}{6}\)}

Vì:  |2\(x\) - 1| = |1 - 2\(x\)|

Nên:      |2\(x\) - 1| + |1 - 2\(x\)| = 8

  ⇒        |2\(x\) - 1| + |2\(x\) - 1| = 8

                         2.|2\(x\) - 1| = 8

                            |2\(x\) - 1| = 8:2

                           |2\(x\) - 1| = 4

                         \(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)

                            \(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)

                             \(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}

giúp mình với

Ta có: \(\left\{{}\begin{matrix}\left(2x-3y\right)^{2018}\ge0\forall x,y\\\left(3y-4z\right)^{2020}\ge0\forall y,z\\\left|2x+3y-z-63\right|\ge0\forall x,y,z\end{matrix}\right.\)

\(\Rightarrow\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|\ge0\forall x,y,z\)

Mà: \(\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|=0\)

nên: \(\left\{{}\begin{matrix}2x-3y=0\\3y-4z=0\\2x+3y-z-63=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=3y\\3y=4z\\z=2x+3y-63\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=4z\\3y=4z\\z=4z+4z-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4z:2\\y=4z:3\\z=8z-63\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=4z:3\\-7z=-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=4\cdot9:3=12\\z=9\end{matrix}\right.\)

Vậy \(x=18;y=12;z=9\).

$Toru$