1) Hình như đề bị sai rồi bạn.

Thông thường pt đã cho sẽ là \(\frac{2x}{x-2}-\frac{5}{x-3}=\frac{5}{x^2-5x+6}\)

Ta thấy \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

Nên ĐKXĐ là \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

pt đã cho \(\Leftrightarrow\frac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{2x^2-6x-5x+10}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)\(\Rightarrow2x^2-11x+5=0\)(*)

Ta có \(\Delta=\left(-11\right)^2-4.2.5=81>0\)nên pt (*) có 2 nghiệm phân biệt:

\(\orbr{\begin{cases}x_1=\frac{-\left(-11\right)+\sqrt{81}}{2.2}=5\left(nhận\right)\\x_2=\frac{-\left(-11\right)-\sqrt{81}}{2.2}=\frac{1}{2}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{1}{2};5\right\}\)

2) Nhận thấy \(3x^2-27=3\left(x^2-9\right)=3\left(x-3\right)\left(x+3\right)\)nên ĐKXĐ ở đây là \(x\ne\pm3\)

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

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

\(\Leftrightarrow\frac{1-3x-9}{3x^2-27}=\frac{1}{4}\)\(\Rightarrow-12x-32=3x^2-27\)\(\Leftrightarrow3x^2+12x+5=0\)(#)

Nhận thấy \(\Delta'=6^2-3.5=21>0\)

Vậy pt (#) có 2 nghiệm phân biệt \(\orbr{\begin{cases}x_1=\frac{-12+\sqrt{21}}{3}\left(nhận\right)\\x_2=\frac{-12-\sqrt{21}}{3}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{-12\pm\sqrt{21}}{3}\right\}\)

ĐKXĐ: \(x\ne-2;-3;-4\)

Ta có: \(x+\frac{x}{x+2}+\frac{x+3}{x^2+5x+6}+\frac{x+4}{x^2+6x+8}=1\)

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

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

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

<=>x=0

Vậy x=0

CỘNG 2 VÀO MỖI VẾ

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\text{ma}:\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

=> x + 0 = 10

=> x       = 0 -10

=> x       = -10

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

Cộng cả 2 vế với 2 ta được :

\(\Leftrightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\times\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà : \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

\(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne4\end{cases}}\)

\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)

\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)

\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)

\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)

\(\Leftrightarrow-6x+1=0\)

\(\Rightarrow-6x=-1\)

\(\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

\(a,\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\left(x\ne1;x\ne-2\right)\)

\(\Leftrightarrow\frac{3}{x^2+x-2}-\frac{1}{x-1}+\frac{7}{x+2}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{6x-8}{\left(x-1\right)\left(x+2\right)}=0\)

=> 6x-8=0

<=> x=\(\frac{8}{6}=\frac{4}{3}\left(tmđk\right)\)

b) ĐKXĐ: x khác 2; x khác 4

\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)

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

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

<=> x^3 - 7x^2 + 16x - 12 = -x^3 + x^2 + 8x - 12

<=> x^2 - 7x^2 + 16x - 12 + x^3 - x^2 + 8x - 12 = 0

<=> 2x^3 - 8x^2 + 8x = 0

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

<=> 2x = 0 hoặc x - 2 = 0

<=> x = 0 (tmđk) hoặc x = 2 (ktmđk)

=> x = 2

TK MÌNH ĐI RỒI MÌNH GIẢI CHO.

cau a: 8x^3 -12x^2 + 6x + 1 =29

<=>8x^3 - 12x^2 + 6x - 28 =0

<=>(8x^3 - 16x^2)+(4x^2 - 8x)+(14x-28)=0

<=>8x^2 ( x-2) + 4x(x-2) + 14(x-2)=0

<=>(x-2)(8x^2 + 4x +14)=0

<=>8x^2 +4x +14 =0 <=> 8(x^2 +1/2 x +7/4)=0<=>(x^2 +2* x*1/4  + 1/16) +27/16 =0 <=>(x+ 1/4)^2=-27/16 (0xay ra) (loai)

=>(x-2)(8x^2 +4x+14)=0 <=> x-2=0 <=>x=2

Vay tap nghiem phuong trinh S={2}

b) x+3/x+2 +x+6/x+5=x+5/x+4 + x+4/x+3

=(x+3/x+2 -1) +(x+6/x+5 - 1)=(x+5/x+4 -1) +(x+4/x+3 -1)

 =1/x+2 +1/x+5 -1/x+4 -1/x+3 =0

x=1000

Bạn chú ý cách viết phương trình.

Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.

\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)

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

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

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

\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)

\(=16\)

Phương trình đã cho trở thành

\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)