a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

Bài làm

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

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

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

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

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

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

b) Bạn kiểm tra lại đề bài

c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)

\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)

\(\Leftrightarrow\frac{-3}{4x+1}-\frac{2}{4x+1}+\frac{8}{\left(4x+1\right)\left(4x-1\right)}=0\)

\(\Leftrightarrow\frac{-3\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3}{\left(4x-1\right)\left(4x+1\right)}-\frac{8x-2}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3-8x+2+8}{\left(4x-1\right)\left(4x+1\right)}=0\)

=> -20x+13=0

<=> -20x=-13

<=> \(x=\frac{13}{20}\left(tmđk\right)\)

\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\)                              \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)

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

\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)

\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)

Vậy pt vô nghiệm

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

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

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

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

ĐKXĐ: x \(\ne\) 1 và x \(\ne\) - 3

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

3x² + 9x - x - 3 - 2x² + 2x + 5x - 5 + 4 = x² - x + 3x - 3

3x² + 9x - x - 3 - 2x² + 2x + 5x - 5 + 4 - x² + x - 3x + 3 = 0

13x - 1 = 0

x = \(\frac{1}{13}\)

chính là 1/13 

nếu đúng thì

tae tae ơi khó quá hổng hiểu j hết trơn

mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được 

\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0 

<=> x + 59 = 0 

<=> x=-59