a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

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

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

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

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

y'=0

=>\(\left(x+3\right)^2-4=0\)

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

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

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

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

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

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

\(\rightarrow\dfrac{1}{2}x+\dfrac{1}{2}+\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{3}x-\dfrac{2}{3}\)

\(\rightarrow\dfrac{1}{2}x+\dfrac{1}{4}x+\dfrac{1}{3}x=3-\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{3}{4}\)

\(\rightarrow\dfrac{13}{12}x=\dfrac{13}{12}\)

\(\rightarrow x=1\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+1\right)+14=0\)

\(\Leftrightarrow x^2-4x+3-\left(x^2-x-2\right)+14=0\)

\(\Leftrightarrow x^2-4x+17-x^2+x+2=0\)

=>-3x+19=0

hay x=19/3(nhận)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(\dfrac{x-1}{x+1}-\dfrac{x-2}{x-3}+\dfrac{14}{x^2-2x-3}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}+\dfrac{14}{\left(x+1\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)-\left(x+1\right)\left(x-2\right)+14}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Rightarrow\left(x^2-4x+3\right)-\left(x^2-x-2\right)+14=0\\ \Leftrightarrow x^2-4x+3-x^2+x+2+14=0\)

\(\Leftrightarrow-3x+19=0\\ \Leftrightarrow x=\dfrac{19}{3}\left(tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{19}{3}\right\}\)

\(\dfrac{x+1}{x-1}-\dfrac{x-2}{x-3}=3\) (ĐK: \(x\ne1;x\ne-3\))

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

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

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

\(\Leftrightarrow7x+1=3\left(x^2-x+3x-3\right)\)

\(\Leftrightarrow3x^2+6x-9-7x-1=0\)

\(\Leftrightarrow3x^2-x-10=0\)

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

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

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

ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{x\left(3-x\right)}{x+1}\cdot\left(x+\dfrac{3-x}{x+1}\right)=2\)

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

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

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

\(\Leftrightarrow4x\left(3-x\right)=2\left(x+1\right)^2\)

\(\Leftrightarrow12x-4x^2=2\left(x^2+2x+1\right)\)

\(\Leftrightarrow-4x^2+12x=2x^2+4x+2\)

\(\Leftrightarrow-4x^2+12x-2x^2-4x-2=0\)

\(\Leftrightarrow-6x^2+8x-2=0\)

\(\Leftrightarrow-6x^2+6x+2x-2=0\)

\(\Leftrightarrow-6x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-6x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-6x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-6x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;\dfrac{1}{3}\right\}\)

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

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

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

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

\(\Leftrightarrow\)\(\dfrac{3x^3+9x-x^4-3x^2}{x^2+2x+1}=2\)

\(\Rightarrow3x^3+9x-x^4-3x^2=2x^2+4x+2\)

\(\Leftrightarrow3x^3+9x-x^4-3x^2-2x^2-4x-2=0\)

\(\Leftrightarrow3x^3-x^4+5x-5x^2-2=0\)

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

Vì x^2 -x+2 >0

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

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy phương trình có nghiệm duy nhất là x=1

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

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

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

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

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

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

ĐKXĐ: \(x\ne1\)

\(x^3+\left(\dfrac{x}{x-1}\right)^3+\dfrac{3x^2}{x-1}-2=0\)

\(\Leftrightarrow\left(x+\dfrac{x}{x-1}\right)^3-3x.\dfrac{x}{x-1}\left(x+\dfrac{x}{x-1}\right)+\dfrac{3x^2}{x-1}-2=0\)

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

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

\(\Leftrightarrow\dfrac{x^2}{x-1}-1=1\)

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

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

Pt đã cho vô nghiệm

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

=>3x-3+2=2x-1

=>3x-1=2x-1

hay x=0

b: (x+1)(x-3)=0

=>x+1=0 hoặc x-3=0

=>x=-1 hoặc x=3

c: \(\Leftrightarrow x\left(x-1\right)-\left(2x-3\right)\left(x+1\right)=x+3\)

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

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

=>x=0(nhận) hoặc x=-1(loại)

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)