1) \(3x\left(x-4\right)-x+4=0\)

\(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

2) \(2x\left(2x+3\right)-2x-3=0\)

\(\Rightarrow2x\left(2x+3\right)-\left(2x+3\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(3x\left(x-4\right)-x+4=0\\ \Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ 2x\left(2x+3\right)-2x-3=0\\ \Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a)-28+x=-34+(-11)             b)-12(3+x)=0

<=>-28+x=-45                   <=>-36-12x=0

<=>x=-17                            <=>-12x=36

Vậy x=-17                             <=>x=-3

                                         Vậy x=-3

c)(7-x)(-x+2)=0

<=>7-x=0 hoặc -x+2=0

Th1:7-x=0                    Th2:-x+2=0

 <=>x=7                             <=>x=2

           Vậy xϵ{7;2}

d)16x2-64=0

<=>32x=64

<=>x=2

Vậy x=2

x-4-√x-2=0(x\(\ge\)2,x-4\(\ge\)\(\sqrt{ }\)x-2)

<=>x-4=√x-2

<=>(x-4)^2=x-2

<=>x^2-8x+16=x-2

<=>x^2-8x-x+16+2=0

<=>x^2-9x+18=0

có △=(-9)^2-4.18=9>0

=>x1=(9+√9)/2=6(thỏa mãn)

x2=(9-√9)/2=3(loại)(vì 3-4=-1,-1<1)

=>x=6 

điều kiện ấy tui nghĩ là (x≥4) sửa lại hộ

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

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)

\(\Leftrightarrow-2x^2-10x=0\)

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

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

Vậy \(S=\left\{0;5\right\}\)

\(x^3+3x^2+3x=0\\ \Leftrightarrow x\left(x^2+3x+3\right)=0\\ \Leftrightarrow x=0\left(x^2+3x+3=x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\right)\)

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

\(\Rightarrow x\left(x^2+3x+3\right)=0\)

Mà: \(x^2+3x+3>0\)

=> x = 0

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

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

mà \(x^2+3x+3=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x=0\)