a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)

\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)

\(\Rightarrow64x+96=72x-24\)

\(\Rightarrow8x=120\Rightarrow x=15\)

b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)

\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)

\(\Rightarrow221x-34=152x+380\)

\(\Rightarrow69x=414\Rightarrow x=6\)

\(\dfrac{2x+1}{3x+2}=\dfrac{x-1}{x-2}\) (đk: x≠ 2; \(-\dfrac{2}{3}\) )

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

⇔ \(2x^2+x-4x-2=3x^2+2x-3x-2\)

⇔ \(3x^2-x-2-2x^2+3x+2=0\)

⇔ \(x^2+2x=0\)

⇔ \(x\left(x+2\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;-2\right\}\)

\(\Leftrightarrow3x^2-3x+2x-2=2x^2-4x+x-2\)

\(\Leftrightarrow x^2+2x=0\)

=>x(x+2)=0

=>x=0 hoặc x=-2

\(Đk:\) \(x\ne1,x\ne2,x\ne3\)

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

\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Rightarrow0x-14=x-10\)

\(\Rightarrow x=-4\left(tmđk\right)\)

=>7/-(1+2x)=5/2x-3

=>7/2x-1=5/2x-3

=>7.(2x-3)=5.(2x-1)

=>14x-21=10x-5

=>10x-(14x-21)=5

=>10x-14x+21=5

=>10x-14x=5-21

=>(-4)x=-16

=>x=4

" . "là nhân ,còn " / "là phân số

ok nha bạn

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

-7.(2\(x\) - 3) = 5.(1 + 2\(x\))

-14\(x\) + 21 = 5 + 10\(x\) 

10\(x\) + 14\(x\) = 21 - 5

24\(x\)           = 16

    \(x\)           =  16 : 24

    \(x\)          =   \(\dfrac{2}{3}\)

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