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Lời giải:
Ta sử dụng các hằng đẳng thức đáng nhớ, cụ thể là công thức:
\((a-b)(a+b)=a^2-b^2\)
a)
\(2003.2005=(2004-1)(2004+1)=2004^2-1^2=2004^2-1< 2004^2\)
Vậy \(2003.2005< 2004^2\)
b)
\(8(7^8+1)(7^4+1)(7^2+1)=(7+1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}.(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^2-1)(7^2+1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^4-1)(7^4+1)(7^8+1)\)
\(=\frac{1}{6}(7^8-1)(7^8+1)=\frac{1}{6}(7^{16}-1)< 7^{16}-1\)
Ta có: \(8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)=\frac{1}{6}.48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^8-1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^{16}-1\right)\)
Vì \(7^{16}-1>\frac{1}{6}\left(7^{16}-1\right)\) nên \(7^{16}-1>8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)\)
1.
a. -3a - 1 + 1 > -3b - 1 + 1 (cộng cả 2 vế cho 1)
-3a . \(\left(\dfrac{-1}{3}\right)\) < -3b . \(\left(\dfrac{-1}{3}\right)\) (nhân cả vế cho \(\dfrac{-1}{3}\) )
a < b
b. 4a + 3 + (- 3) < 4b + 3 +(- 3) (cộng cả 2 vế cho -3)
4a . \(\dfrac{1}{4}\) < 4b . \(\dfrac{1}{4}\) (nhân cả 2 vế cho \(\dfrac{1}{4}\) )
a < b
2.
a. Ta có: a < b
3a < 3b ( nhân cả 2 vế cho 3)
3a - 7 < 3b - 7 (cộng cả 2 vế cho - 7 )
b. Ta có: a < b
-2a > -2b (nhân cả 2 vế cho -2)
5 - 2a > 5 - 2b ( cộng cẩ 2 vế cho 5)
c. Ta có: a < b
2a < 2b (nhân cả vế cho 2)
2a + 3 < 2b + 3 (cộng cả 2 vế cho 3)
d. Ta có: a < b
3a < 3b (nhân cả 2 vế cho 3)
3a - 4 < 3b - 4 (cộng cả 2 vế cho -4)
Ta có: 3 < 4
đến đây ko bắt cầu qua đc chắc đề bài sai
\(a< b\)
\(\Leftrightarrow-2a>-2b\)
\(\Leftrightarrow7-2a>7-2b\)
\(\Leftrightarrow5\left(7-2a\right)>5\left(7-2b\right)\)
\(\Leftrightarrow5\left(7-2a\right)-8>5\left(7-2b\right)-8\)
\(35-10a-8=27-10a\)
\(35-10b-8=27-10b\)
a<b ==> 27-10a > 27-10b
==> 5(7-2a) > 5(7-2b)-8
\(D=8\left(7^8+1\right)\left(7^4+1\right)\left(7^2-1\right)\)
\(D=\frac{4}{25}\left(7^2+1\right)\left(7^2-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^8-1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^{16}-1\right)\)
Vì: \(\frac{4}{25}\left(7^{16}-1\right)< 7^{16}-1\Rightarrow D< C\)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(a,2003\cdot2005=\left(2004-1\right)\left(2004+1\right)=2004^2-1< 2004^2\)
\(b,7^{16}-1\\ =\left(7^8-1\right)\left(7^8+1\right)=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =\left(7-1\right)\left(7+1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)>8\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
a. Dựa vào tính chất thừa và thiếu, suy ra: 2003 . 2005 = 20042