Ta có \(\frac{7x-5}{8}=\frac{3x+1}{5}\)

<=> \(35x-25=24x+8\)

<=> \(35x-24x=8+25\)

<=> \(11x=33\)

<=>  x=3

Vậy phương trình có ngiệm duy nhất x=3

\(\frac{7x-5}{8}=\frac{3x+1}{5}\)                         MTC= 40

\(5\left(7x-5\right)=8\left(3x+1\right)\)

\(35x-25=24x+8\)

\(35x-24x=8+25\)

\(11x=33\)

\(\Rightarrow x=3\)

(3x + 5)² - (2x + 1)² = 0

<=> (3x + 5 + 2x + 1)(3x + 5 - 2x - 1) = 0

<=> (5x + 6)(x + 4) = 0

<=> \(\orbr{\begin{cases}x=-\frac{6}{5}\\x=-4\end{cases}}\)

Vậy \(x\in\left\{-\frac{6}{5};-4\right\}\)là nghiệm phương trình

\(\left(3x+5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(3x+5+2x+1\right)\left(3x+5-2x-1\right)=0\)

\(\Leftrightarrow\left(5x+6\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=-\frac{6}{5}\)

Vậy tập nghiệm của phương trình là S = { -4 ; -6/5 } 

\(\frac{5x-3}{6}-\frac{7x-1}{4}-\frac{4x+2}{7}+5=0\)

<=> \(\frac{14\left(5x-3\right)-21\left(7x-1\right)-12\left(4x+2\right)+420}{84}=0\)

<=> 70x - 42 - 147x + 21 - 48x -24 + 420 = 0

<=> -125x + 375 = 0

<=> -125x = -375

<=> x = 3

Vậy S = {3}

\(\frac{3\left(2x+1\right)}{4}-5-\frac{3x+2}{10}=\frac{2\left(3x-1\right)}{5}\)

<=> \(\frac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\frac{8\left(3x-1\right)}{20}\)

<=> 30x + 15 - 100 - 6x - 4 = 24x - 8

<=> 24x - 24x = -8 + 89

<=> 0x = 81

=> pt vô nghiệm

\(a,4x-6< 7x-12\)

\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)

\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)

\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)

\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)

\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)

\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)

\(\Leftrightarrow-16x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{16}\)

\(d,-12-8x>3+2x-\left(5-7x\right)\)

\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)

\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)

\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)

\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)

\(b, (2x^2 + 3x-1) - 5(2x^2 + 3x + 2) + 24 =0 \)

Đặt \(2x^2 + 3x + 1 = a \)

\(=> (a-2) - 5(a+2) + 24 = 0\)\(\)

\(=> a - 2 - 5a - 10 + 24 = 0\)

\(=> a = 3=> 2x^2 + 3x + 1 = 3\)

\(<=> 2x^2 + 3x - 2 = 0\)

\(<=> 2x^2 + 4x - x - 2 = 0\)

\(<=> (2x-1)(x+2) = 0 \)

\(<=> 2x - 1 = 0 hoặc x+2 =0\)

\(<=> x = 1/2 hoặc x = -2\)

~~

\(\frac{3x^2+7x-10}{x}=0\)

\(3x^2+7x-10=0\)

\(3x^2-3x+10x-10=0\)

\(3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\left(3x+10\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+10=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-10}{3}\\x=1\end{cases}}\)

\(ĐKXĐ:\)\(x\ne0\)

       \(\frac{3x^2+7x-10}{x}=0\)

\(\Rightarrow\)\(3x^2+7x-10=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\3x+10=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\left(TMĐKXĐ\right)\\x=-\frac{10}{3}\left(TMĐKXĐ\right)\end{cases}}\)

Vậy...

\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!