a)$\frac{x-2}{x+3}$ = $\frac{x-3}{x+1}$ b)$\frac{x}{3}$ = $\frac{y}{4}$ và xy=48 c)2,7:(2x)=$\frac{11}{5}$ :3

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1. Đáp án:

   $\text{ a) $\dfrac{x-2}{x+3}$ = $\dfrac{x-3}{x+1}$ }$

   $\text{ĐKXĐ : x $\neq$ -3 ; x $\neq$ -1}$

   $\text{⇔ $\dfrac{(x-2)(x+1)}{(x+3)(x+1)}$ = $\dfrac{(x-3)(x+3)}{(x+1)(x+3)}$ }$

   $⇒ (x-2)(x+1) = (x-3)(x+3)$

   $⇔ x² +x -2x -2 = x² – 3x +3x – 9$

   $⇔ x² – x² +x -2x-3x+3x = -9 +2 $

   $⇔ -x = -7$

   $⇔ x= 7(Thỏa mãn)$

   $\text{Vậy x= 7 }$

   $\text{b) $\dfrac{x}{3}$ = $\dfrac{y}{4}$ và xy = 48}$

   $\text{Đặt $\dfrac{x}{3}$ = $\dfrac{y}{4}$ = k}$

   $⇒ x = 3k$

   $⇒ y = 4k$

   $\text{Theo đề , ta có :  }$

   $ 3k . 4k = 48$

   $⇔ 12k² = 48$

   $⇔ k² = 48 : 12$

   $⇔ k² = 4$

   $⇔k=±2$

   $\text{Thay k =2 vào, ta có :}$

   $⇒x = 2 .3 = 6$

   $⇒y = 2 . 4 = 8$

   $\text{Thay k = -2 vào, ta có :}$

   $⇒ x = -2 . 3 = -6$

   $⇒y = -2 . 4 = -8$

   $\text{Vậy các cặp số x,y cần tìm là}$ $ (-6 ; -8) , (6 ; 8)$

   $c) 2,7 : (2x) = \dfrac{11}{5} : 3$

   $⇔ 2,7 : (2x) = \dfrac{11}{15}$

   $⇔ 2x = 2,7 : \dfrac{11}{15}$

   $⇔ 2x= \dfrac{81}{22}$

   $⇔ x= \dfrac{81}{22} : 2$

   $⇔ x= \dfrac{81}{44}$

   $\text{Vậy x = $\dfrac{81}{44}$ }$

    

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2. a) $\dfrac{x-2}{x+3}=\dfrac{x-3}{x+1}$

   ⇒ $(x-2)(x+1)=(x+3)(x-3)$

   ⇒ $[(x-2)x]+(x-2)=x^2-3^2$

   ⇒ $x^2-2x+x-2=x^2-9$

   ⇒ $x^2+(-2x+x)-2=x^2-9$

   ⇒ $x^2-x-2=x^2-9$

   ⇒ $(x^2-x-2)-(x^2-9)=0$

       $x^2-x-2-x^2+9=0$

       $(x^2-x^2)-x+(-2+9)=0$

       $-x+7=0$

   ⇒ $-x=-7$

   ⇒ $x=7$

   b) $\dfrac{x}{3}=\dfrac{y}{4}$

   ⇒ $x=\dfrac{3y}{4}$

   ⇒ $xy=\dfrac{3y}{4}.y=48$

   ⇒ $xy=\dfrac{3y^2}{4}=48$

       $xy=3y^2=48.4=192$

       $y^2=192:3=64$

   \(⇒\left[ \begin{array}{l}y=8\\x=6\end{array} \right.\)

   \(⇒\left[ \begin{array}{l}y=-8\\x=-6\end{array} \right.\)

   c) $2,7:(2x)=\dfrac{11}{5}:3$

   $\dfrac{27}{10}:(2x)=\dfrac{11}{15}$

   $2x=\dfrac{27}{10}:\dfrac{11}{15}$

   $2x=\dfrac{27}{10}.\dfrac{15}{11}=\dfrac{81}{22}$

   ⇒ $x=\dfrac{81}{22}:2$

   ⇒ $x=\dfrac{81}{22}.\dfrac{1}{2}=\dfrac{81}{44}$

    

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