x - 3 / x -2   -  x - 2 /x -4  =16/5

x - 3 / x - 2   -  x - 2 /x -4   - 16/5  = 0

-16^2 +81x -88/ 5(x-2)(x-4) = 0

-16^2 +81x -81 =0

16^2 -81x +88 =0

x = -(-81) ± √(-81)^2 -4 *16 *88 /2*16

x = 81±√ 929/32

x1 =81+√929/32

x-2 =81-√929/32

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

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

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

\(\Rightarrow\frac{x^2-7x+12+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)

.................

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

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

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

\(\Rightarrow\left(x^3-1\right)\left[2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)\right]=\left(x^3-1\right)\left(2x-1\right)\left(2x+1\right)\)

\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\)

\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)

\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-\left(4x^2-1\right)=0\)

\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0\)

\(\Rightarrow3x=0\)

\(\Rightarrow luon-dung-voi-moi-x\)

nhầm phải là

3x=0

=>không có giá trị x thỏa mãn yêu cầu

\(x\ne2;4\)

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

\(\Leftrightarrow x^2-7x+12+x^2-4x+4+x^2-6x+8=0\)

\(\Leftrightarrow3x^2-17x+24=0\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{8}{3}\end{matrix}\right.\)

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

Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:

\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)

\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)

\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)

\(\Leftrightarrow2t^2+t-1=6t^2-6t\)

\(\Leftrightarrow-4t^2+7t-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)

Vậy phương trình vô nghiệm.

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-4x+4-3x-6=2x-22\)

\(\Leftrightarrow x^2-7x-2-2x+22=0\)

\(\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

8,

b, (-x²+12x+4)/(x²+3x-4) = 12/(x+4) + 12/(3x-3)

(=) (-x²+12x+4)/(x-1)(x+4) -12(x-1)/(x-1)(x+4) - 4(x+4)/(x-1)(x+4) = 0

(=) -x² +12x +4 -12x +12 -4x -16 = 0

(=) -x² -4x = 0

(=) -x(x+4) = 0

(=) -x = 0 hoặc x +4 = 0

(=) x=0 hoặc x=-4

Vậy S={0;4}

Chúc bạn học tốt.

\(1.\frac{7x-3}{x-1}=\frac{2}{3}\)   ( \(x\ne1\))

\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)

\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)

\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)

\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-5\)

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

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

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

\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)

\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)

\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)

\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)

\(\Leftrightarrow4x^2+5x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)

\(\left(2x+\frac{5}{4}\right)^2>0\)

\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)

=> PT vô nghiệm 

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

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

\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)

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

\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

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

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

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

\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{16}{6}\)

\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)

\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)

\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)

\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)

\(\Leftrightarrow x^4+x^3-4x-8=0\)

\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)

Đến đấy mk tắc r xl bạn nhé