a.

\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)

\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)=18\)

Đặt \(t=x^2+2x+1=\left(x+1\right)^2\left(t\ge0\right)\)

\(\Rightarrow\left(4t-1\right)\cdot t=18\)

\(\Leftrightarrow\left(2t\right)^2-2\cdot2t\cdot\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{289}{16}\)

\(\Leftrightarrow\left(2t-\dfrac{1}{4}\right)^2=\dfrac{289}{16}\Leftrightarrow\left(t-\dfrac{1}{8}\right)^2=\dfrac{289}{64}\)

\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{8}=\dfrac{17}{8}\\t-\dfrac{1}{8}=-\dfrac{17}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\\t=-2\left(loai\right)\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^2=\dfrac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x+1=\dfrac{3}{2}\\x+1=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{5}{2};\dfrac{1}{2}\right\}\)

b.

Ta có:

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

- \(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)

- \(x^2+18x+80=x^2+8x+10x+80=x\left(x+8\right)+10\left(x+8\right)=\left(x+8\right)\left(x+10\right)\)

Thay vào phương trình, ta được:

\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{x+10-\left(x+1\right)}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\Leftrightarrow\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=25\)

\(\Leftrightarrow x^2+11x+\dfrac{121}{4}=\dfrac{181}{4}\)

\(\Leftrightarrow\left(x+\dfrac{11}{2}\right)^2=\dfrac{181}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{11}{2}=\dfrac{\sqrt{181}}{2}\\x+\dfrac{11}{2}=-\dfrac{\sqrt{181}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{181}}{2}\\x=\dfrac{-11-\sqrt{181}}{2}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{-11+\sqrt{181}}{2};\dfrac{-11-\sqrt{181}}{2}\right\}\)

c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)

=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12

=>-16x-1=-15x+12

=>-x=13

=>x=-13

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>9x^2-23x=x^2-5x+6

=>8x^2-18x-6=0

=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)

b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>216x+18=275x-100

=>-59x=-118

=>x=2

ĐKXĐ: \(x\ne\left\{-10;-8;-3;-1\right\}\)

\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\)

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)

Suy ra: \(3x-6-x+1=2x+4\)

\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)

2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)

Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)

\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)

\(\Leftrightarrow5x=16\)

hay \(x=\dfrac{16}{5}\)

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

\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)

`<=> 20(2x-1) +15(3x-2) =12(4x-3)`

`<=> 40x - 20 + 45x - 30 = 48x - 36`

`<=> 85x -50 = 48x - 36`

`<=> 85x-48x = -36+50`

`<=> 37x =14`

`<=> x= 14/37`

Vậy phương trình có nghiệm `x=14/37`

__

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)

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

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

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

`=> 5x + 15 + 4x -12=x-6`

`<=> 9x + 3=x-6`

`<=> 9x-x=-6-3`

`<=> 8x = -9`

`<=>x=-9/8(tm)`

Vậy phương trình có nghiệm `x=-9/8`

` @ yngoc`

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)

\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)

\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)