1)500-{5.[409-(2^2.3-21)^2]}:15 2)3x-31=14 3)20+5.(x-8)=5^2.4 4)A=2^0+2^1+2^2+…………….+2^49+2^50

Bởi

1)500-{5.[409-(2^2.3-21)^2]}:15
2)3x-31=14
3)20+5.(x-8)=5^2.4
4)A=2^0+2^1+2^2+…………….+2^49+2^50

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  1. Đáp án:

    \(\begin{array}{l}
    1)\,\,\frac{{1172}}{3}\\
    2)\,\,x\, = 15\\
    3)\,\,x = 24\\
    4)\,\,\,A = {2^{51}} – 1.
    \end{array}\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    1)\,\,\,500 – \left\{ {5.\left[ {409 – {{\left( {{2^2}.3 – 21} \right)}^2}} \right]} \right\}:15\\
    = 500 – \left\{ {5.\left[ {409 – {{\left( {12 – 21} \right)}^2}} \right]} \right\}:15\\
    = 500 – 5.\left( {409 – {9^2}} \right):15\\
    = 500 – 5.\left( {409 – 81} \right):15\\
    = 500 – 5.328:15\\
    = 500 – \frac{{328}}{3}\\
    = \frac{{500.3 – 328}}{3} = \frac{{1172}}{3}.\\
    2)\,\,3x – 31 = 14\\
    \,\,\,\,\,\,3x\,\,\,\, = 14 + 31\\
    \,\,\,\,\,\,3x\,\,\,\,\,\, = 45\\
    \,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\, = 45:3\\
    \,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\, = 15.\\
    3)\,\,\,20 + 5.\left( {x – 8} \right) = {5^2}.4\\
    \,\,\,\,\,\,\,\,20 + 5\left( {x – 8} \right) = 100\\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\left( {x – 8} \right) = 80\\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x – 8 = 16\\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\, = 24.\\
    4)\,\,\,A = {2^0} + {2^1} + {2^2} + ……. + {2^{49}} + {2^{50}}\\
    Ta\,\,\,co:\,\,\,\,2A = 2\left( {{2^0} + {2^1} + {2^2} + ……. + {2^{49}} + {2^{50}}} \right)\\
    \Rightarrow 2A = {2^1} + {2^2} + {2^3} + …… + {2^{50}} + {2^{51}}\\
    \Rightarrow 2A – A = {2^1} + {2^2} + {2^3} + …… + {2^{50}} + {2^{51}} – \left( {{2^0} + {2^1} + {2^2} + ……. + {2^{49}} + {2^{50}}} \right)\\
    \Rightarrow A = {2^{51}} – {2^0}\\
    \Rightarrow A = {2^{51}} – 1.
    \end{array}\)

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