1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)

\(=-x^2+2x+5x-10+x^2-49=7x-59\)

\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)

\(=9x^2+6x+1-9x^2+4=6x+5\)

=>7x-59=6x+5

=>x=64

2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)

\(=5x^2+5x-x-1-2x^2+12x-9\)

\(=3x^2+16x-10\)

\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)

\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)

\(=3x^2-4x-18\)

=>16x-10=-4x-18

=>20x=-8

hay x=-2/5

\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

a: =-3/4-1/4+2/7+5/7+2023/2024

=-1+1+2023/2024=2023/2024

b: 2/3x=2/7

=>x=2/7:2/3=3/7

c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5*3/2=9/10

vt rõ đề câu a đi lm 1 thể

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{3}{5}\)\(x\) - 1

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{3}{5}\)\(x\) =  - 1 + \(\dfrac{1}{3}\)

 (\(\dfrac{2}{7}\) - \(\dfrac{3}{5}\))\(x\) =  - \(\dfrac{2}{3}\)

  - \(\dfrac{11}{35}\)\(x\)     = - \(\dfrac{2}{3}\)

         \(x\)     = - \(\dfrac{2}{3}\) : (- \(\dfrac{11}{35}\))

         \(x\)     = \(\dfrac{70}{33}\)

Vậy \(x=\dfrac{70}{33}\) 

|3\(x\) + 7| - 3\(x\) = 7

|3\(x\) + 7| = 7  +3\(x\) Vì 7 + 3\(x\) > 0  nên 3\(x\) >  -7 ⇒ \(x\) >  -\(\dfrac{7}{3}\)

Với \(x\) >  - \(\dfrac{7}{3}\) ta có:

3\(x\) + 7 = 7 + 3\(x\) 

        7  =7 (luôn đúng)

Vậy \(x\) \(\ge\) - \(\dfrac{7}{3}\)