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

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

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

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

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

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

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

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

đánh đề bằng latex cho rõ đi bạn, không biết nào dấu nào biến:v

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

\(x-\dfrac{1}{2}=\dfrac{3}{5}\times6\)

\(x-\dfrac{1}{2}=\dfrac{18}{5}\)

\(x=\dfrac{18}{5}+\dfrac{1}{2}\)

\(x=\dfrac{41}{10}\)

d: \(\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}\)

\(\dfrac{1}{3-x}=\dfrac{-1}{x-3}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=8\\x+1=-8\end{matrix}\right.\Leftrightarrow x\in\left\{7;-9\right\}\)

\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)

\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)

\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)

\(\Leftrightarrow x=2\)

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

ĐK:\(x\ge0\)

\(\left(x^2-1\right)\sqrt{x}=0\Leftrightarrow\left(x-1\right)\left(x+1\right)\sqrt{x}=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\\x=0\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

Ủa lớp 7 sao học căn r nè

Đề có sai không em ơi !