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a) Ta có F = \(\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
=> 8F = \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^8-1\right)\left(3^8+1\right)-3^{16}=3^{16}-1-3^{16}=-1\)
=> F = -1/8
b) Ta có G = \(\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)
=> 7G = 7(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (23 - 1)(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (26 - 1)(26 + 1)(212 + 1) - 224
=> 7G = (212 - 1)(212 + 1) - 224
=> 7G = 224 - 1 - 224
=> 7G = -1
=> G = -1/7
\(F=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
<=> \(\left(3^2-1\right)F=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\left(3^2-1\right)\frac{3^{16}}{8}\)
<=> \(8F=\left(3^4-1\right)\left(3^4+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^8+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^{16}-1\right)-3^{16}=-1\)
<=> F = -1/8
Câu G làm tương tự
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)+1\)
Đặt x = \(2^2+1\)ta được :
\(3x.x^2.x^4.x^8\)+1
= \(3x^{15}\)+1
thay x=\(2^2+1\)ta được:
\(3\left(2^2+1\right)^{15}+1=3.5^{15}+1\)
3(22+1)(24+1)(28+1)(216+1)+1
=3x5x17x257x65537+1
=4294967296
chọn đúng cho mình điểm nha!
Áp dụng HĐT đáng nhớ :
\(\left(a-b\right)\left(a+b\right)=a^2-b^2\) . Ta có :
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
Chúc bạn học tốt !!!
Bài 1:
a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)
b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(18^8-\left(18^8-1\right)=1\)
\(c,100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)
áp dụng công thức Gauss ta đc đáp án là:10100
d, mk khỏi ghi đề dài dòng:
\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:
\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)
\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)
\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)
1c,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1 \)