Giải :

\(\frac{\left(3x-1\right)(x+2)}{3}-\frac{2x^2+1}{2}=\frac{11}{2}\Leftrightarrow\frac{2\left(3x-1\right)\left(x+2\right)-3\left(2x^2+1\right)}{6}=\frac{33}{6}\)

                                                                           \(\Leftrightarrow2\left(3x-1\right)\left(x+2\right)-3\left(2x^2+1\right)=33\)

                                                                           \(\Leftrightarrow\left(6x^2+10x-4\right)-\left(6x^2+3\right)=33\)

                                                                           \(\Leftrightarrow6x^2+10x-4-6x^2-3=33\)

                                                                           \(\Leftrightarrow10x=33+4+3\)

                                                                           \(\Leftrightarrow10x=40\)

                                                                           \(\Leftrightarrow x=4\)

Phương trình có tập nghiệm \(S=\left\{4\right\}\).

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

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

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

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

<=> 4 = 3 (vô lý)

=> pt vô nghiệm

b) ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)

<=> 2x - x² - 4 + 2x - 5x - 5x² + 10 = 15x - 30

<=> -x + 4x² - 14 = 15x - 30

<=> x - 4x² + 14 = 15x - 30 

<=> x - 4x² + 14 + 15x - 30 = 0

<=> 16x - 4x² - 16 = 0

<=> 4(4x - x² - 4) = 0

<=> -x² + 4x - 4 = 0

<=> x² - 4x + 4 = 0

<=> (x - 2)² = 0

<=> x - 2 = 0

<=> x = 2 (ktm)

=> pt vô nghiệm 

c) xem bài 4 ở đây: Câu hỏi của gjfkm

d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)

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

<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)

<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)

<=> x² - 3x + 4x - 12 + x² - 2x + x - 2 = 2x² - 4x + 5x - 10

<=> 2x² - 14 = 2x² + x - 10

<=> 2x² - 14 - 2x² = x - 10

<=> -14 = x - 10

<=> -14 + 10 = x

<=> -4 = x

<=> x = -4

\(\text{GIẢI :}\)

ĐKXĐ : \(x\ne\pm1\)

\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)

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

\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}-\frac{x+1}{x^2-1}=0\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow\text{ }2\left(x-1\right)+x\left(x+1\right)-(x+1)=0\)

\(\Leftrightarrow\text{ }2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1\text{ (loại)}\\x=-3\text{ (Chọn)}\end{cases}}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-3\right\}\).

\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}.\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)\(đk:x\ne\pm1\)

\(< =>\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left[\frac{7}{6}.\frac{6}{7}+\left(1\right)\right]x+1}{x^2-1}\)

\(< =>\frac{2x-2+x^2+x}{x^2+x-x-1}=\frac{2x+1}{x^2-1}\)\(< =>\frac{x^2+3x-2}{x^2-1}=\frac{2x-1}{x^2-1}\)

\(< =>x^2+2x-2=2x-1\)\(< =>x^2+2x-2x-2+1=0\)

\(< =>x^2-1=0< =>x^2=1\)\(< =>x=\pm1\)\(\left(ktmđk\right)\)

Vậy phương trình trên vô nghiệm

ĐKXĐ: \(x\ne\pm2\)

\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\left(tm\right)\)

Vậy: \(S=\left\{-\frac{7}{23}\right\}\)

=.= hk tốt!!

Giải :

\(\text{ĐKXĐ}\: :\: x\ne\pm2\)

\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

 \(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

Khử mẫu : \(\left(-6x^2-12x+x+2\right)+\left(9x^2-18x+4x-8\right)=3x^2-2x+1\)

           \(\Leftrightarrow-23x=7\Leftrightarrow x=\frac{7}{23}\).

giúp mik vs

ĐKXĐ:\(x\ne1\)

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

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

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

\(\Leftrightarrow x^2+3x-1=3x^2\)\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(KTMĐK\right)\\x=\frac{1}{2}\left(TMĐK\right)\end{cases}}}\)

Vậy nghiệm của pt là \(x=\frac{1}{2}\)

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

\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)

đến đây là pt bậc 2