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a) Ta có : x(x + 4)(x - 4) - (x2 + 1)(x2 - 1)
= x(x2 - 16) - (x4 - 1)
= x3 - 16x - x4 + 1
= x(x2 - 16 - x3) + 1
\(a,x.\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-16x=x^4+1\)
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
\(B=2^{32}\)
=> \(A< B\)
ta có A= \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=(2-1)(2+1)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(2^{32}-1\) (ấp dụng các hằng đẳng thức )
=> A=232-1
B=232
=> A<B
A= \(\frac{3\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{\left(2^2-1\right)}=2^{32-1}\)
mà B= \(2^{32}\)
=> A<B
tách ít ít ra thôi. để cả cộp thế này k ai làm cho đâu. mệt quá
B=\(2^{16}-1\)
\(A=2+1.2^2+1.2^4+1.2^8+1\)\(=\left(2.2^2.2^4.2^8\right)+\left(1+1+1+1\right)\)\(=2^{15}+4\)
mà \(2^{16}>2^{15}\)=> A>B
b) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(=2^{64}-1-2^{64}=-1\)
Ta có:
a) A = 2018 x 2020 = (2019 - 1) x (2019 + 1)
Áp dụng hằng đẳng thức thứ ba ta có:
A = 208 x 2020 = \(2019^2-1^2=2019^2-1\)
Vì \(2019^2-1< 2019^2\)
\(\Rightarrow\)A < B
b) A = \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1^2\right)\left(2^2+1^2\right)\left(2^4+1^2\right)\left(2^8+1^2\right)\left(2^{16}+1^2\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Vì \(2^{32}-1< 2^{32}\)
\(\Rightarrow\)A < B
a) Áp dụng hàng đăng thức (a - b) (a + b) = a2 - b2
Ta có : A = 2018.2020 = (2019 - 1) (2019 + 1) = 20192 - 1
Mà B = 20192
Nên A < B