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1. Đáp án:

     bài 3:

   `a)102+2(x-3)=108`

   `⇔2(x-3)=108-102`

   `⇔2(x-3)=6`

   `⇔x-3=3`

   `⇔x=6`

   `b) [(x^2+2)+17]:5=20`

   `⇔x^2+2+17=20.5`

   `⇔x^2+19=100`

   `⇔x^2=100-19`

   `⇔x^2=81`

   `⇔x=+-9`

   `c) (x-1)^4=81`

   Th1: `(x-1)^4=3^4⇔x-1=3⇔x=4`

   Th2: `(x-1)^4=(-3)^4⇔x-1=-3⇔x=-2`

   bài 4:

   `A=1+4+4^2+4^3+…+4^100`

   `4A=4+4^2+4^3+…+4^101`

   `4A-A=(4+4^2+…+4^101)-(1+4+4^2+…+4^100)`

   `3A=4^101-1`

   `A=(4^101-1)/3`

   khi đó:

   `B=(4^101-1)/3-A=(4^101-1)/3-(4^101-1)/3=0`

   vậy `B=0`

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2. bài 3:

   a) $102+2(x-3)=108$

   $↔2(x-3)=108-102=6$

   $↔x-3=6:2=3$

   $↔x=3+3=6$

   b) $[(x^2+2)+17]:5=20$

   $↔(x^2+2)+17=20.5=100$

   $↔x^2+2=100-17=83$

   $↔x^2=83-2=81$

   \(↔\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\)

   c) $(x-1)^4=81$

   \(↔\left[ \begin{array}{l}x-1=3\\x-1=-3\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=4\\x=-2\end{array} \right.\)

   Bài 4:

   $A=1+4+4^2+4^3+…+4^{100}$

   $4A=4+4^2+4^3+…+4^{101}$

   $4A-A=(4+4^2+4^3+…+4^{101})-(1+4+4^2+…+4^{100}$

   $3A=4^{101}-1$

   $→A=\dfrac{4^{101}-1}{3}$

   $→B=(4^{101}-1):3-A=\dfrac{4^{101}-1}{3}-\dfrac{4^{101}-1}{3}=0$

   Reply