1 – (x+4) + (x+8) + (x+12) + … + (x+132) = 5577 – x^2020 = x^10 – 3^x+2 + 3^x = 2430 17/07/2021 Bởi Everleigh 1 – (x+4) + (x+8) + (x+12) + … + (x+132) = 5577 – x^2020 = x^10 – 3^x+2 + 3^x = 2430
Đáp án: $\begin{array}{l}a)\left( {x + 4} \right) + \left( {x + 8} \right) + \left( {x + 12} \right) + … + \left( {x + 132} \right) = 5577\\ \Rightarrow \left( {x + x + .. + x} \right) + \left( {4 + 8 + 12 + … + 132} \right) = 5577\end{array}$ Có số số hạng là:$\dfrac{{132 – 4}}{4} + 1 = 33$ $\begin{array}{l} \Rightarrow 33.x + \dfrac{{\left( {132 + 4} \right).33}}{2} = 5577\\ \Rightarrow 33.x + 2244 = 5577\\ \Rightarrow 33.x = 3333\\ \Rightarrow x = 101\end{array}$ Vậy x=101 $\begin{array}{l}b){x^{2020}} = {x^{10}}\\ \Rightarrow {x^{2020}} – {x^{10}} = 0\\ \Rightarrow {x^{10 + 2010}} – {x^{10}} = 0\\ \Rightarrow {x^{10}}.{x^{2010}} – {x^{10}} = 0\\ \Rightarrow {x^{10}}.\left( {{x^{2010}} – 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}{x^{10}} = 0\\{x^{2010}} = 1\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 0\\x = 1\\x = – 1\end{array} \right.\end{array}$ Vậy x=0;x=1;x=-1 $\begin{array}{l}c){3^{x + 2}} + {3^x} = 2430\\ \Rightarrow {3^x}{.3^2} + {3^x} = 2430\\ \Rightarrow {3^x}.\left( {{3^2} + 1} \right) = 2430\\ \Rightarrow {3^x}.10 = 2430\\ \Rightarrow {3^x} = 243\\ \Rightarrow {3^x} = {3^5}\\ \Rightarrow x = 5\end{array}$ Vậy x=5 Bình luận
Đáp án:
$\begin{array}{l}
a)\left( {x + 4} \right) + \left( {x + 8} \right) + \left( {x + 12} \right) + … + \left( {x + 132} \right) = 5577\\
\Rightarrow \left( {x + x + .. + x} \right) + \left( {4 + 8 + 12 + … + 132} \right) = 5577
\end{array}$
Có số số hạng là:$\dfrac{{132 – 4}}{4} + 1 = 33$
$\begin{array}{l}
\Rightarrow 33.x + \dfrac{{\left( {132 + 4} \right).33}}{2} = 5577\\
\Rightarrow 33.x + 2244 = 5577\\
\Rightarrow 33.x = 3333\\
\Rightarrow x = 101
\end{array}$
Vậy x=101
$\begin{array}{l}
b){x^{2020}} = {x^{10}}\\
\Rightarrow {x^{2020}} – {x^{10}} = 0\\
\Rightarrow {x^{10 + 2010}} – {x^{10}} = 0\\
\Rightarrow {x^{10}}.{x^{2010}} – {x^{10}} = 0\\
\Rightarrow {x^{10}}.\left( {{x^{2010}} – 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^{10}} = 0\\
{x^{2010}} = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1\\
x = – 1
\end{array} \right.
\end{array}$
Vậy x=0;x=1;x=-1
$\begin{array}{l}
c){3^{x + 2}} + {3^x} = 2430\\
\Rightarrow {3^x}{.3^2} + {3^x} = 2430\\
\Rightarrow {3^x}.\left( {{3^2} + 1} \right) = 2430\\
\Rightarrow {3^x}.10 = 2430\\
\Rightarrow {3^x} = 243\\
\Rightarrow {3^x} = {3^5}\\
\Rightarrow x = 5
\end{array}$
Vậy x=5