Giải pt :< $\frac{2}{(x-4)(x-2)}$ +$\frac{x+3}{x-4}$=$\frac{x+1}{2-x}$ 14/07/2021 Bởi Eden Giải pt :< $\frac{2}{(x-4)(x-2)}$ +$\frac{x+3}{x-4}$=$\frac{x+1}{2-x}$
` 2/((x-4)(x-2)) +(x+3)/(x-4)=(x+1)/(2-x)` ĐK : ` x \ne 2;\ x \ne 4` `\to 2/((x-4)(x-2)) + ( (x+3)(x-2))/((x-4)(x-2)) = – ((x+1)(x-4))/((x-4)(x-2))` `\to ( 2 + (x+3)(x-2) + (x+1)(x-4))/((x-4)(x-2)) = 0` `\to 2 + (x^2 + x -6) + (x^2 – 3x -4) =0` `\to 2x^2 – 2x -8 = 0` `\to x^2 – x -4 = 0` ` \to x^2 – x +1/4 -17/4 = 0` `\to (x-1/2)^2 =17/4` `\to` \(\left[ \begin{array}{l}x – \dfrac{1}{2} = \dfrac{\sqrt{17}}{2}\\\\x – \dfrac{1}{2} = \dfrac{-\sqrt{17}}{2}\end{array} \right.\) `\to` \(\left[ \begin{array}{l}x = \dfrac{\sqrt{17}+1}{2}\\\\x = \dfrac{-\sqrt{17}+1}{2}\end{array} \right.\) Vậy ` x \in { \frac{-\sqrt{17}+1}{2}; \frac{\sqrt{17}+1}{2}}` Bình luận
` 2/((x-4)(x-2)) +(x+3)/(x-4)=(x+1)/(2-x)` ĐK : ` x \ne 2;\ x \ne 4`
`\to 2/((x-4)(x-2)) + ( (x+3)(x-2))/((x-4)(x-2)) = – ((x+1)(x-4))/((x-4)(x-2))`
`\to ( 2 + (x+3)(x-2) + (x+1)(x-4))/((x-4)(x-2)) = 0`
`\to 2 + (x^2 + x -6) + (x^2 – 3x -4) =0`
`\to 2x^2 – 2x -8 = 0`
`\to x^2 – x -4 = 0`
` \to x^2 – x +1/4 -17/4 = 0`
`\to (x-1/2)^2 =17/4`
`\to` \(\left[ \begin{array}{l}x – \dfrac{1}{2} = \dfrac{\sqrt{17}}{2}\\\\x – \dfrac{1}{2} = \dfrac{-\sqrt{17}}{2}\end{array} \right.\)
`\to` \(\left[ \begin{array}{l}x = \dfrac{\sqrt{17}+1}{2}\\\\x = \dfrac{-\sqrt{17}+1}{2}\end{array} \right.\)
Vậy ` x \in { \frac{-\sqrt{17}+1}{2}; \frac{\sqrt{17}+1}{2}}`
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