Toán Lớp 8: tìm x
(3x-5)^2- 4(x-1)^2=0
rút gọn biểu thức
D=3(2^2-1)(2^4+1)…….. (2^64+1)+1
gấp
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Toán Lớp 8: tìm x
(3x-5)^2- 4(x-1)^2=0
rút gọn biểu thức
D=3(2^2-1)(2^4+1)…….. (2^64+1)+1
gấp
Comments ( 1 )
{\left( {3x – 5} \right)^2} – 4{\left( {x – 1} \right)^2} = 0\\
\Leftrightarrow {\left( {3x – 5} \right)^2} – {\left( {2x – 2} \right)^2} = 0\\
\Leftrightarrow \left( {3x – 5 – 2x + 2} \right)\left( {3x – 5 + 2x – 2} \right) = 0\\
\Leftrightarrow \left( {x – 3} \right)\left( {5x – 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 3 = 0\\
5x – 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \frac{7}{5}
\end{array} \right.\\
Vay\,x = 3;x = \frac{7}{5}\\
D = 3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)….\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^2} – 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)….\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)….\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^8} – 1} \right).\left( {{2^8} + 1} \right)…\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^{64}} – 1} \right)\left( {{2^{64}} + 1} \right) + 1\\
= {2^{128}} – 1 + 1\\
= {2^{128}}
\end{array}$