câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

a: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

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

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

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

\(pt\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)

\(\Leftrightarrow32x^4+48x^3+16x^2+48x^3+72x^2+24x+18x^2+27x+9-810=0\)

\(\Leftrightarrow32x^4+96x^3+106x^2+51x-801=0\)

\(\Leftrightarrow32x^4+96x^3+106x^2+318x-267x-801=0\)

\(\Leftrightarrow\left(x+3\right)\left(32x^3+106x-267\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(16x^2+24x+189\right)=0\)

Vì \(16x^2+24x+89=\left(4x+3\right)^2+80\ge80\) nên \(\orbr{\begin{cases}x+3=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)

Ta có: \(\left(4x+3\right)^2\left(x+1\right)\left(2x+1\right)=810\)

\(\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)

Đặt \(a=2x^2+3x+1\)

\(\Rightarrow\left(8a+1\right)a=810\)

\(\Leftrightarrow8a^2+a-810=0\)

\(\Leftrightarrow\left(a-10\right)\left(8a+81\right)=0\)

\(\Rightarrow\left(2x^2+3x-9\right)\left(16x^2+24x+189\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(16x^2+24x+189\right)=0\)

Lại có: \(16x^2+24x+189=\left(4x+3\right)^2+80>0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)

\(4+2x\left(2x+4\right)=-x\)

\(4+2x.2x+8x=-x\)

\(4x+8x+x=-4\)

\(13x=-4\)

\(x=-\frac{4}{13}\)

 Vậy pt có nghiệm là { -4/13 }

2) mình nghĩ thế này

(2x-3)^2=2x-3

Đẻ 2 cái trên = nhau thfi 

2x-3=1

=> x=2

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

\(\Leftrightarrow21\left(x+13\right)=7\left(2x-1\right)-3\left(5x+2\right)\)

\(\Leftrightarrow21x+273=14x-7-15x-6=-x-13\)

=>22x=-286

hay x=-13

b: \(\dfrac{2x-3}{3}-\dfrac{x-3}{6}=\dfrac{4x+3}{5}-17\)

\(\Leftrightarrow10\left(2x-3\right)-5\left(x-3\right)=6\left(4x+3\right)-510\)

\(\Leftrightarrow20x-30-5x+15=24x+18-510\)

\(\Leftrightarrow15x-15=24x-492\)

=>-9x=-477

hay x=53