Tìm x: a, $\frac{x+10}{9}$ = $\frac{x-5}{7}$ b, $\frac{x+8}{20}$ = $\frac{5}{x+8}$ c, $\frac{x+4}{5}$ = $\frac{x-8}{6}$ d, $\frac{x-2}{9}$ =

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1. $a$) $\dfrac{x+10}{9} = \dfrac{x-5}{7}$

   $⇔ 7(x+10) = 9(x-5)$

   $⇔7x + 70 = 9x – 45$

   $⇔ 70 + 45 = 9x-7x$

   $⇔ 2x = 115$

   $⇔ x = \dfrac{115}{2}$

     Vậy $x = \dfrac{115}{2}$

   $b$) $\dfrac{x+8}{20} = \dfrac{5}{x+8}$

   $⇔ (x+8)^2 = 100$

   $⇔ x+8 = ±10$

   $⇒$ \(\left[ \begin{array}{l}x=2\\x=-18\end{array} \right.\) 

     Vậy $x$ $∈$ `{-18;2}`

   $c$) $\dfrac{x+4}{5} = \dfrac{x-8}{6}$

   $⇔ 6(x+4) = 5(x-8)$

   $⇔ 6x + 24 = 5x – 40$

   $⇔ 24+40 = 5x-6x$

   $⇔ -x = 64$

   $⇔ x = -64$ 

     Vậy $x=-64$

   $d$) $\dfrac{x-2}{9} = \dfrac{4}{x-2}$

   $⇔ (x-2)^2 = 36$

   $⇒ x-2 = ± 6$

   $⇒$ \(\left[ \begin{array}{l}x=8\\x=-4\end{array} \right.\) 

      Vậy $x$ $∈$ `{-4;8}`

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2. Đáp án:

   \(a,\ x=\dfrac{115}{2}\\ b,\ x=2\ \text{hoặc}\ x=(-18)\\ c,\ x=(-64)\\ d,\ x=8\ \text{hoặc}\ x=(-4)\)

   Giải thích các bước giải:

   \(a,\ \dfrac{x+10}{9}=\dfrac{x-5}{7}\\ ⇔7(x+10)=9(x-5)\\ ⇔7x+70=9x-45\\ ⇔7x-9x=-45-70\\ ⇔-2x=-115\\ ⇔x=\dfrac{-115}{-2}=\dfrac{115}{2}\\ \text{Vậy x = $\dfrac{115}{2}$}\\ b,\ \dfrac{x+8}{20}=\dfrac{5}{x+8}\\ ⇔(x+8)^{2}=5.20\\ ⇔(x+8)^{2}=100\\ ⇔(x+8)^{2}=(±10)^{2}\\ ⇔\left[ \begin{array}{l}x+8=10\\x+8=-10\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=2\\x=-18\end{array} \right.\\ \text{Vậy x = 2 hoặc x = (-18)}\\ c,\ \dfrac{x+4}{5}=\dfrac{x-8}{6}\\ ⇔6(x+4)=5(x-8)\\ ⇔6x+24=5x-40\\ ⇔6x-5x=-40-24\\ ⇔x=-64\\ \text{Vậy x = (-64)}\\ d,\ \dfrac{x-2}{9}=\dfrac{4}{x-2}\\ ⇔(x-2)^{2}=9.4\\ ⇔(x-2)^{2}=36\\ ⇔(x-2)^{2}=(±6)^{2}\\ ⇔\left[ \begin{array}{l}x-2=6\\x-2=-6\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=8\\x=-4\end{array} \right.\\ \text{Vậy x = 8 hoặc x = (-4)}\)

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