\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\) (đk: x \(\ne2;4\))
<=> \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}+1=0\)
<=> \(\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2+\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=0\)
<=> \(x^2-7x+12+x^2-4x+4+x^2-6x+8=0\)
<=> \(3x^2-17x+24=0\)
<=> (x-3)(3x-8) = 0
<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)
KL: x \(\in\left\{\dfrac{8}{3};3\right\}\)
\(\dfrac{x-3}{x-2}\) + \(\dfrac{x-2}{x-4}\) = -1
ĐKXĐ : x - 2 ≠ 0
\(\Rightarrow\) x ≠ 2
x - 4 ≠ 0
\(\Rightarrow\) x ≠ 4
MTC : (x - 2) (x - 4)
Quy đồng mẫu thức hai vế của phương trình
\(\Rightarrow\) \(\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\) + \(\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\) = - \(\dfrac{\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
Suy ra : (x - 3) (x - 4) + (x - 2) (x - 2) = - (x - 2) (x - 4)
\(\Leftrightarrow\) x2 - 7x + 12 + x2 - 4x + 4 = - (x2 - 6x + 8)
\(\Leftrightarrow\) x2 - 7x + 12 + x2 - 4x + 4 = -x2 + 6x - 8
\(\Leftrightarrow\) x2 - 7x + 12 + x2 - 4x + 4 + x2 - 6x + 8 = 0
\(\Leftrightarrow\) 3x2 - 17x + 24 = 0
\(\Leftrightarrow\) 3x2 -9x - 8x + 24 = 0
\(\Leftrightarrow\) (3x2 - 9x) - (8x + 24)
\(\Leftrightarrow\) 3x(x - 3) - 8(x - 3) = 0
\(\Leftrightarrow\) (x - 3) (3x - 8) = 0
Khi x - 3 = 0 hoặc 3x - 8 = 0
\(\Leftrightarrow\) x = 3(thỏa mãn) \(\Leftrightarrow\) 3x = 8
\(\Leftrightarrow\) x = \(\dfrac{8}{3}\) (thỏa mãn)
Vậy S = \(\left\{3,\dfrac{8}{3}\right\}\)
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