Bạn ơi bạn có thể viết rõ hơn được ko
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
\(\Leftrightarrow\) \(\frac{4\left(2x^2-6x+1\right)}{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}-\frac{3\left(x^2-3x+2\right)}{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}+\frac{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}=0\)
\(\Leftrightarrow\) 4(2x2 - 6x + 1) - 3(x2 - 3x + 2) + (x2 - 3x + 2)(2x2 - 6x + 1) = 0
\(\Leftrightarrow\) 8x2 − 24x + 4 − 3x2 + 9x − 6 + 2x4 − 12x3+ 23x2 − 15x + 2 = 0
\(\Leftrightarrow\) 2x4 − 12x3 + 28x2 − 30x = 0
\(\Leftrightarrow\) 2x(x3 - 6x2 + 14x - 5) = 0
\(\Leftrightarrow\) 2x(x3 - 3x2 - 3x2 + 9x + 5x - 15) = 0
\(\Leftrightarrow\) 2x(x2 - 3x + 5)(x - 3) = 0
\(\Leftrightarrow\) 2x = 0 hoặc x2 - 3x + 5 = 0 hoặc x - 3 = 0
Ta xét x2 - 3x + 5
\(\Leftrightarrow\) (x - \(\frac{3}{2}\))2 + \(\frac{11}{4}\)
Ta có: (x - \(\frac{3}{2}\))2 \(\ge\) 0 với mọi x
\(\Rightarrow\) (x - \(\frac{3}{2}\))2 + \(\frac{11}{4}\) \(\ge\) \(\frac{11}{4}\) > 0 với mọi x
hay x2 - 3x + 5 > 0 với mọi x
\(\Rightarrow\) 2x = 0 hoặc x - 3 = 0
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy S = {0; 3}
Chúc bạn học tốt!
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}=0\)
\(\frac{4}{\left(x-2\right)\left(x-1\right)}-\frac{3}{2x^2-6x+1}=0\)
\(28x^2-30x+2x^4-12x^3=0\)
\(2x\left(x^2-3x+5\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}x=0\\x^2-3x+5=0\left(voli\right)\\x=3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
