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Ta có 1999*2001 = (2000-1)*(2000+1)
= 2000^2 - 1^2
Biết 2000^2 = 2000^2
=> 2000^2 - 1^2 < 2000^2
<=> 1999*2001 < 2000^2
a)72 -272 =(7-27)(7+27)
=-20.34
=-680
b)372 -132=(37-13)(37+13)=24.50
=1200
c)20022-22=(2002-2)(2002+2)
=2000.2004
=4008000
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
a) \(x^2+6xy+9x^2=\left(x+3x\right)^2\)
b) \(\left(a-2b^2\right)^2=a^2-4ab^2+4b^4\)
c) \(\left(m+1\right)^2=m^2+2m+1\)
d) \(m^2-4n^4=\left(m+2n^2\right)\left(m-2n^2\right)\)
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
\(B=2^{32}\)
=> \(A< B\)
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
\(=\left\{\left(2x+2\right)+\left(2x-2\right)\right\}^2\)2
\(=\left(4x\right)^2\)
\(=16x^2\)
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
A= 216 và B=(2+1)(22+1)(24+1)(28+1)
Xet B=(2+1)(22+1)(24+1)(28+1)
=(2-1)(2+1)(22+1)(24+1)(28+1)
=(22-1)(22+1)(24+1)(28+1)
=(24-1)(24+1)(28+1)
=(28-1)(28+1)
=216-1
So sanh A=216 va B=216-1 ta co A>B