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

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

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

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

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

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

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

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

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

\(\left(2x-1\right)^2=81\\ \Rightarrow\left(2x-1\right)=\sqrt{81}=9\\ \Rightarrow2x=9+1=10\\ \Rightarrow x=\dfrac{10}{2}=5\)

\(2^3.5-\left(x+3^2\right)=10\\ \Rightarrow8.5-\left(x+9\right)=10\\ \Rightarrow x+9=8.5-10=40-10=30\\ \Rightarrow x=30-9=21\)

a: P(x)=0

=>4x-7-x-14=0

=>3x-21=0

=>x=7

b: x^2+x=0

=>x(x+1)=0

=>x=0; x=-1

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x=14

x=14:7=2

ta có: 3/x - 5 = -4/x + 2

     => 3(x + 2) = -4(x + 5)

     => 3x + 6 = (-4)x + -20

     => 3x + 6 = (-4)x - 20

     => (-4)x - 3x = 20 + 6

     => (-1)x = 26

     => x = -26

\(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}+.......+\frac{100}{x}=101\)

\(\Leftrightarrow\frac{1}{x}\left(1+2+3+....+100\right)=101\)

\(\Leftrightarrow\frac{1}{x}.\frac{100.\left(100+1\right)}{2}=101\)

\(\Leftrightarrow\frac{1}{x}.50=1\)

\(\Rightarrow\frac{1}{x}=\frac{1}{50}\)

\(\Rightarrow x=50\)

cảm ơn bạn nhiều nhé Hùng .

(-1)+3+(-5)+7+....+x = 600

<=> [(-1) + 3] + [(-5) + 7] .... + [(-x) - 2) + x] = 600 

Ta co : 2 + 2 + 2 +.....+ 2 = 600 

<=> 1 + 1 + 1 +.....+ 1 = 300 

Số dấu ngoặc[ ] la : x−34 +1

=> x−34 +1=300

<=> x−34 =299

<=> x - 3 = 299 . 4 = 1199

Vậy x = 1199

bn ui khi nào cần xong

bn tách số âm thành 1 vế, số dương thành 1 vế rồi tính như tính dãy số có quy luật nha

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

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

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

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

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