\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\\ \Rightarrow7.\left(2x-1\right)-3.\left(5x+2\right)=21.\left(x+13\right)\\ \Rightarrow14x-7-15x-6=21x+273\\\Rightarrow -x-21x=273+13\\ \Rightarrow-22x=286\\ \Rightarrow x=-13\\ b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}=0\\ \Rightarrow9.\left(x+3\right)+6=4.\left(5x+9\right)-3.\left(7x-9\right)=0\\\Rightarrow 9x+27+6=20x+36-21x+27\\ \Rightarrow9x+33=-x+63\\ \Rightarrow10x=30\\ \Rightarrow x=3\)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Rightarrow7\left(2x-1\right)-3\left(5x+2\right)-21x-273=0\)

\(\Rightarrow14x-7-15x-6-21x-273=0\)

\(\Rightarrow-22x=286\)

\(\Rightarrow x=-13\)

\(b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}\)

\(\Rightarrow9\left(x+3\right)+6-4\left(5x+9\right)+3\left(7x-9\right)=0\)

\(\Rightarrow9x+27+6-20x-36+21x-27=0\)

\(\Rightarrow10x=30\Rightarrow x=3\)

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+2-\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-\dfrac{7}{2}x+2\right)=0\)

=>x=1 hoặc x=4/7

e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

=>10x-4=15-9x

=>19x=19

hay x=1

f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

=>2x-1+x-1=1

=>3x-2=1

hay x=1(loại)

g: =>1+3x-6=3-x

=>3x-5-3+x=0

=>4x-8=0

=>x=2(loại)

c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)

\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)

e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)

\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)

a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)

\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)

hay \(x=-\dfrac{41}{12}\)

Vậy: \(x=-\dfrac{41}{12}\)

giải nhanh giup mình nhé

\(a,ĐK:5x\ge0\Leftrightarrow x\ge0\\ b,ĐK:3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\\ c,ĐK:5-x\ge0\Leftrightarrow x\le5\\ đ,ĐK:3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\\ f,ĐK:\dfrac{-3}{1+4x}\ge0\Leftrightarrow1+4x< 0\left(-3< 0;1+4x\ne0\right)\\ \Leftrightarrow x< -\dfrac{1}{4}\\ h,ĐK:10+x^2\ge0\Leftrightarrow x\in R\left(10+x^2\ge10>0\right)\)

b.

\(3x\left(x-2\right)=5x-10\)

\(\Leftrightarrow3x^2-6x=5x-10\)

\(\Leftrightarrow3x^2-6x-5x+10=0\)

\(\Leftrightarrow\left(3x^2-6x\right)-\left(5x-10\right)=0\)

\(\Leftrightarrow3x\left(x-2\right)-5\left(x-2\right)=0\)

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

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