Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :

\(\frac{x+4}{2x^2-5x+2}-\frac{x+4}{2x^2-7x+3}=0\)

\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{2x^2-5x+2}-\frac{1}{2x^2-7x+3}\right)=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)

\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)

Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)

Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)

Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)

ttiiok

a,\(2x\left(x-3\right)=x-3.\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy ..... 

b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow x^2+2x-5x+10=8\)

\(\Leftrightarrow x^2-3x+10-8=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy ....

\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{\left(2x+1\right)\cdot3}{4\cdot3}-\frac{\left(x-5\right)\cdot4}{3\cdot4}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{6x+3}{12}-\frac{4x-20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{6x+3-4x+20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23-4x+1}{12}< 12\)

\(\Leftrightarrow\frac{-2x+24}{12}< 12\)

\(\Leftrightarrow-2x+24< 144\)

\(\Leftrightarrow-2x< 120\)

\(\Leftrightarrow x< -60\)

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

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

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

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

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

Vậy pt vô nghiệm

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

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

ĐK: x≠1

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

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

<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

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

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x²-2x+1)=0

<=> -(6x²-x+12x-2)+9x²+4x-18x-8-3x²+2x-1 = 0

<=> -6x²-11x+2+9x²+4x-18x-8-3x²+2x-1=0

<=> -23x-7=0

<=> -23x=7

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

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

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

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

⇒x²+x+1+2x²-5=4x-4

⇔3x²-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}