A) (x^2-4)(x+3/5)=0 B) x-7/12x+3/8x=5/24

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1. a) $(x^{2} – 4)(x+\frac{3}{5})=0$

   $<=>$ \(\left[ \begin{array}{l}x^{2} – 4=0\\x+\frac{3}{5}=0\end{array} \right.\)

   $<=>$ \(\left[ \begin{array}{l}x^{2} = 4\\x=\frac{-3}{5}\end{array} \right.\)

   $<=>$ \(\left[ \begin{array}{l}x= ±2\\x=\frac{-3}{5}\end{array} \right.\)

   Vậy x∈{2;-2; $\frac{-3}{5}$}

   b) $x-\frac{7}{12}x+\frac{3}{8}x = $ $\frac{5}{24}$ 

   $<=> $ $\frac{19}{24}x$ = $\frac{5}{24}$ 

   $<=>$ $x = \frac{5}{19}$ 

   Vậy $x = \frac{5}{19}$

    

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2. Đáp án + Giải thích các bước giải:

   `A//`

   `(x^{2}-4)(x+(3)/(5))=0`

   `=>` \(\left[ \begin{array}{l}x^{2}-4=0\\x+\dfrac{3}{5}=0\end{array} \right.\) 

   `=>` \(\left[ \begin{array}{l}x^{2}=4\\x=-\dfrac{3}{5}\end{array} \right.\) 

   `=>` \(\left[ \begin{array}{l}x=±2\\x=-\dfrac{3}{5}\end{array} \right.\) 

   Vậy `x∈{±2;-(3)/(5)}`

   `B//`

   `x-(7)/(12)x+(3)/(8)x=(5)/(24)`

   `=>x.1-(7)/(12)x+(3)/(8)x=(5)/(24)`

   `=>x(1-(7)/(12)+(3)/(8))=(5)/(24)`

   `=>x((24)/(24)-(14)/(24)+(9)/(24))=(5)/(24)`

   `=>x.(19)/(24)=(5)/(24)`

   `=>x=(5)/(24):(19)/(24)`

   `=>x=(5)/(24).(24)/(19)`

   `=>x=(5)/(19)`

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