\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

mình cần gấp nha 

bạn làm như bình thường.mà giá trị tuyệt đói thì nhớ làm 2 th nha

study well

 các bạn giúp mình lên 100 sp nha

a: Ta có: \(A=\left|x-3.5\right|+4.1-x\)

\(=x-3.5+4.1-x\)

=0.6

\(\left|x-3,5\right|+\left|y-1,3\right|=0\)

Ta có:

\(\left\{{}\begin{matrix}\left|x-3,5\right|\ge0\\\left|y-1,3\right|\ge0\end{matrix}\right.\forall xy.\)

\(\Rightarrow\left|x-3,5\right|+\left|y-1,3\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3,5=0\\y-1,3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0+3,5\\y=0+1,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\y=1,3\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{3,5;1,3\right\}.\)

Chúc bạn học tốt!

Các bạn giúp mk nhanh vs nhá, chiều mai mk cần. cảm ơn trước

Ta có |x - 3,5| + |y - 1,3| = 0

Mà : |x - 3,5| \(\ge0\forall x\in R\)

       |y - 1,3| \(\ge0\forall x\in R\)

Suy ra : |x - 3,5| = |y - 1,3| = 0

\(\Leftrightarrow\orbr{\begin{cases}x-3,5=0\\y-1,3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3,5\\y=1,3\end{cases}}\)

1)\(3,5\cdot x+1,54\cdot3,5=15\)

    \(3,5\left(x+1,54\right)=15\)

                 \(x+1,54=\frac{15}{3,5}\)

                 \(x+1,54=\frac{30}{7}\)

                                \(x=\frac{30}{7}-1,54\)

                                 \(x=\frac{30}{7}-\frac{77}{50}\)

                               \(x=\frac{961}{350}\)

\(20+\frac{12}{x-2019}=32\)

             \(\frac{12}{x-2019}=32-20\)

              \(\frac{12}{x-2019}=12\)

               \(x-2019=\frac{12}{12}\)

              \(x-2019=1\)

                               \(x=1+2019\)

2)

\(\frac{3}{7}\cdot\frac{4}{13}+\frac{3}{7}\cdot\frac{9}{13}+5\frac{4}{7}\)

\(=\frac{3}{7}\cdot\frac{4}{13}+\frac{3}{7}\cdot\frac{9}{13}+13\cdot\frac{3}{7}\)

\(=\frac{3}{7}\left(\frac{4}{13}+\frac{9}{13}+13\right)\)

\(=\frac{3}{7}\cdot14\)

\(=6\)

3.5*(x-1.54)=15

x+1.54=15;3.5

x+1.54=4.28

x=4.28-1.54

x=2.74

chưa suy nghĩ