a: =>(x-2)(3x+1)-(x-2)(x+2)=0

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

=>(x-2)(2x-1)=0

=>x=1/2 hoặc x=2

b: =>3(x-1)+4(x+1)=6(x-1)

=>3x-3+4x+4=6x-6

=>7x+1=6x-6

=>x=-7

c: =>x(x-3)-(x+2)(x+3)+16=0

=>x^2-3x-x^2-5x-6+16=0

=>10-8x=0

=>x=5/4

a: =>x-2+2=x^2+2x

=>x^2+2x=x

=>x^2+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: =>-9(5x-8)+4(7x-12)=-6(x+18)

=>-45x+72+28x-48=-6x-108

=>-17x+24=-6x-108

=>-11x=-132

=>x=12

\(a.2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\\ \Leftrightarrow\frac{10x}{5}+\frac{6}{5}=\frac{25}{5}-\frac{13+5x}{5}\\\Leftrightarrow 10x+6=25-13-5x\\\Leftrightarrow 15x=6\\ \Leftrightarrow x=\frac{2}{5}\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{2}{5}\right\}\)

Lần sau ghi đề nhớ ghi rõ, đọc đề bạn mình muốn khóc luôn.

\(b.\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\\\Leftrightarrow \frac{21x}{24}-\frac{120\left(x-9\right)}{24}=\frac{80x+6}{24}\\\Leftrightarrow 21x-120x+1080=80x+6\\ \Leftrightarrow-179x=-1074\\\Leftrightarrow x=6\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{6\right\}\)

ko bt giải 

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)

\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)=0\)

\(\Leftrightarrow x-5=0\) (do \(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\))

\(\Rightarrow x=5\)

ĐKXĐ: \(\left\{{}\begin{matrix}3x+1>=0\\6-x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\x< =6\end{matrix}\right.\)

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

=>\(\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)

=>\(\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+3x^2-15x+x-5=0\)

=>\(\dfrac{3\cdot\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)

=>\(\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)

=>x-5=0

=>x=5(nhận)