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

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

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

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

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

\(\left(x-\dfrac{1}{5}\right)^2+1=3,5\div7\%\)

`=> (x-1/5)^2 + 1 = 3,5 \div 0,07`

`=> (x-1/5)^2 +1=50`

`=> (x-1/5)^2 = 49`

`=> (x-1/5)^2 = (+-7)^2`

`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{5}=7\\x-\dfrac{1}{5}=-7\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=7+\dfrac{1}{5}\\x=-7+\dfrac{1}{5}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{36}{5}\\x=-\dfrac{34}{5}\end{matrix}\right.\)

Vậy, `x={36/5; -34/5}.`

X x 3.5 =0.125:80

X x 3.5=10

X= 10:3.5

X=2.85

$x$  .  $3,5$  $=$  $0,125$  $:$  $80$

$x$  .  $3,5$  $=$  $10$

$x$  $=$  $10$  $:$  $3,5$

$x$  $=$  $2,85$

$#BearChan$

=4,3(4,6+3.5+2,9-1)

=4,3*10=43

\(A=4.3\cdot4.6+3.5\cdot4.3+4.3\cdot2.9-4.3\)

\(A=4.3\cdot4.6+3.5\cdot4.3+4.3-2.9-4.3\cdot1\)

\(A=4.3\cdot\left(4.6+3.5+2.9-1\right)\)

\(A=4.3\cdot10\)

\(A=43\)

\(x\times3,5-x\times2=30\)
\(x\times\left(3,5-2\right)=30\)
\(x\times1,5=30\)
\(x=30:1,5\)
\(x=20\)

\(3,5\times x-2\times x=30\\ \Leftrightarrow x\left(3,5-2\right)-30\\ \Leftrightarrow1,5x=30\\ \Leftrightarrow x=30:1,5\\ \Leftrightarrow x=20\)

Vậy \(x=20\)

X x 3,5 - X x 2 = 30

X x (3.5 - 2)    = 30

X x 1.5            = 30

X                     = 30 : 1.5

X                     = 20

a, \(\left|x-\frac{-3}{2}\right|=\frac{1}{4}\Rightarrow\left|x+\frac{3}{2}\right|=\frac{1}{4}\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=\frac{1}{4}\\x+\frac{3}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{-7}{4}\end{cases}}}\)

b, \(\left|x-3,5\right|-\frac{1}{2}=0\Rightarrow\left|x-3,5\right|=\frac{1}{2}\)  đến đây tương tự a

Bài 3:

Gọi (d): y=ax+b là phương trình đường thẳng cần tìm

Vì (d)//y=3x-1 nên a=3 và b<>-1

=>y=3x+b

Thay x=1 và y=-2 vào y=3x+b, ta được:

b+3=-2

=>b=-2-3=-5

Vậy: (d): y=3x-5

Bài 5:

a: Tọa độ giao điểm của (d1) và (d2) là:

\(\left\{{}\begin{matrix}3x-1=x+3\\y=x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\y=x+3\end{matrix}\right.\)

=>x=2 và y=2+3=5

Thay x=2 và y=5 vào (d3), ta được:

\(2\left(m+1\right)+m-3=5\)

=>2m+2+m-3=5

=>3m-1=5

=>3m=6

=>m=2

b: (d3): y=(m-1)x+m-3

=mx-x+m-3

=m(x+1)-x-3

Tọa độ điểm cố định mà (d3) luôn đi qua là:

\(\left\{{}\begin{matrix}x+1=0\\y=-x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\left(-1\right)-3=1-3=-2\end{matrix}\right.\)