Cho $$\dfrac{6}{1+\dfrac{4(x+3)}{7+x}}$$ Tính: `B=\frac{4A}{7+x}`

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1. Đáp án+Giải thích các bước giải:

    `1+(4(x+3))/(7+x)`

   `=(7+x)/(7+x)+(4(x+3))/(7+x)`

   `=(7+x+4x+12)/(7+x)`

   `=(5x+19)/(7+x)`

   `\to 6:(5x+19)/(7+x)`

   `=6.(7+x)/(5x+19)`

   `\toA=6.(7+x)/(5x+19)`

   `\to 4.A=4.6.(7+x)/(5x+19)`

   `\to 4A=(24(7+x))/(5x+19)`

   `\to (4A)/(7+x)=(24(7+x))/(5x+19):(7+x)`

   `\to B=(24(7+x))/(5x+19).(1)/(7+x)`

   `\to B=24/(5x+19)`

   Vậy `B=24/(5x+19)`

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2. \(1+\dfrac{4(x+3)}{7+x}=\dfrac{7+x+4x+12}{7+x}=\dfrac{5x+19}{7+x}\\→A=\dfrac{6}{\dfrac{5x+19}{7+x}}=\dfrac{6(7+x)}{5x+19}\\B=\dfrac{4A}{7+x}\\↔B=\dfrac{24(7+x)}{5x+19}.\dfrac{1}{7+x}=\dfrac{24(7+x)}{(5x+19)(7+x)}\\=\dfrac{24}{5x+19}(x\ne -\dfrac{19}{5})\)

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