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\(=\left(40^2-31^2\right)-\left(39^2-32^2\right)+\left(38^2-33^2\right)-\left(37^2-34^2\right)+\left(36^2-35^2\right)\)
\(=\left(40-31\right)\left(40+31\right)-\left(39-32\right)\left(39+32\right)+\left(38-33\right)\left(38+33\right)-\)
\(\left(37-34\right)\left(37+34\right)+\left(36-35\right)\left(36+35\right)\)
\(=9.71-7.71+5.71-3.71+1.71\)
\(=\left(9-7+5-3+1\right).71=5.71=355\)
\(40^2-39^2+38^2-37^2+....+32^2-31^2\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(32-31\right)\left(32+31\right)\)
\(=40+39+38+37+....+32+31\)
Số số hạng của dãy trên là: (40-31):1+1= 10 (số)
Tổng trên là: (40+31) x 10 : 2 = 355
Vậy ....
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow2^{64}-1\)
\(2003^2-3^2=\left(2003-3\right)\left(2003+3\right)=2000.2006=4012000\)
\(2003^2-3^2=\left(2003-3\right)\left(2003+3\right)=2000\cdot2006=2\cdot1000\cdot2006=4012000\)