1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)

ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow-7x+3=-4x-15\)

\(\Leftrightarrow-7x+4x=-15-3\)

\(\Leftrightarrow-3x=-18\)

\(\Leftrightarrow x=6\)( tmđk )

Vậy x = 6 là nghiệm của phương trình

2) 2x + 3 < 6 - ( 3 - 4x )

<=> 2x + 3 < 6 - 3 + 4x

<=> 2x - 4x < 6 - 3 - 3

<=> -2x < 0

<=> x > 0

Vậy nghiệm của bất phương trình là x > 0

\(ĐK:x\ge-\dfrac{3}{2}\\ \Leftrightarrow\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\\sqrt{2x-3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\2x-3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\left(đk:x\ge\dfrac{3}{2}\right)\)

\(\Leftrightarrow4x^2-9=4\left(2x+3\right)\)

\(\Leftrightarrow4x^2-9=8x+12\)

\(\Leftrightarrow4x^2-8x-21=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\left(tm\right)\\x=-\dfrac{3}{2}\left(ktm\right)\end{matrix}\right.\)

\(ĐK:x\le-\dfrac{3}{2};\dfrac{3}{2}\le x\\ Pt\Leftrightarrow\sqrt{\left(2x+3\right)\left(2x-3\right)}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\) đk \(x\ge\dfrac{3}{2}\)

\(\Leftrightarrow4x^2-9=4\left(2x+3\right)\)

\(\Leftrightarrow4x^2-9=8x+12\)

\(\Leftrightarrow4x^2-8x-21=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+3\right)=0\)

\(\left[{}\begin{matrix}x=\dfrac{7}{2}\left(nhận\right)\\x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)

Vậy S=\(\left\{\dfrac{7}{2}\right\}\)

\(\Leftrightarrow\left(12x+9\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(12x+9-2x+2\right)\left(12x+9+2x-2\right)=0\)

\(\Leftrightarrow\left(10x+11\right)\left(14x+7\right)=0\)

=>x=-11/10 hoặc x=-1/2

\(9\left(4x+3\right)^2=4\left(x^2-2x+1\right)\\ \Leftrightarrow\left[3\left(4x+3\right)\right]^2-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(12x+9\right)^2-\left[2\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(12x+9\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(12x+9-2x+2\right)\left(12x+9+2x-2\right)=0\\ \Leftrightarrow\left(10x+11\right)\left(14x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{10}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow4x^2-12x+9-8x^2+18=0\)

\(\Leftrightarrow-4x^2-12x+27=0\)

\(\Leftrightarrow4x^2-6x+18x-27=0\)

\(\Leftrightarrow2x\left(2x-3\right)+9\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+9\right)=0\)

\(\left[{}\begin{matrix}3-2x=0\\2x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

Vậy ...

=>(2x-3)(2x+3)(x-4)-(2x-3)(x-4)(x+4)=0

=>(2x-3)(x-4)(2x+3-x-4)=0

=>(2x-3)(x-4)(x-1)=0

=>\(x\in\left\{1;4;\dfrac{3}{2}\right\}\)

c) \(\dfrac{x}{x-2}+\dfrac{x}{x+2}=\dfrac{4x}{x^2-4}.ĐKXĐ:x\ne2;-2\)

<=>\(\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{x\left(x-2\right)}{x^2-4}=\dfrac{4x}{x^2-4}\)

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

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

<=>\(\left[{}\begin{matrix}2x=0< =>x=0\\x-2=0< =>x=2\left(loại\right)\end{matrix}\right.\)

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

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

Vậy pt trên có tập nghiệm là S={\(\dfrac{-3}{2};\dfrac{4}{3}\)}

a) 5x+9 =2x

<=> 5x-2x=9

<=> 3x=9

<=> x=3

Vậy pt trên có nghiệm là S={3}

b) (x+1)(4x-3)=(2x+5)(x+1)

<=> (x+1)(4x-3)-(2x+5)(x+1)=0

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

<=>\(\left[{}\begin{matrix}x+1=0< =>x=-1\\2x-8=0< =>2x=8< =>x=4\end{matrix}\right.\)

Vậy pt trên có tập nghiệm là S={-1;4}

c) 

<=>

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

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

<=>

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

<=> 

Vậy pt trên có tập nghiệm là S={}

\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{5-x}=b\\\sqrt[3]{4x-3}=c\\\sqrt[3]{9-2x}=d\end{matrix}\right.\) 

Ta được: \(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)

TH1:

Nếu \(a+b=c+d=0\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}=0\)

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

TH2: nếu \(a+b=c+d\ne0\)

\(a+b=c+d\Leftrightarrow\left(a+b\right)^3=\left(c+d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3+d^3+3cd\left(c+d\right)\)

\(\Leftrightarrow ab\left(a+b\right)=cd\left(c+d\right)\) (do \(a^3+b^3=c^3+d^3\))

\(\Leftrightarrow ab=cd\) (do \(a+b=c+d\ne0\))

\(\Leftrightarrow\sqrt[3]{\left(3x+1\right)\left(5-x\right)}=\sqrt[3]{\left(4x-3\right)\left(9-2x\right)}\)

\(\Leftrightarrow\left(3x+1\right)\left(5-x\right)=\left(4x-3\right)\left(9-2x\right)\)

\(\Leftrightarrow5x^2-28x+32=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{8}{5}\end{matrix}\right.\)

Vậy \(x=\left\{-3;4;\dfrac{8}{5}\right\}\)

Cái cuối này căn bậc 2 hay căn bậc 3 em? Căn bậc 2 thì hơi nghi ngờ về khả năng giải được của pt này.