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1) A=19952-1994.1996
=19952-(1995-1)(1995+1)
=19952-(19952-1)
=1
2) B=98.28-(184-1)(184+1)
=(9.2)8-[(184)2-1]
= 188-188+1
=1
3) C=1632+74.163+372
=1632+2.37.163+372
=1632+2.163.37+372
=(163+37)2.2
=80000
như thế này chứ:
A=1002-992+982-972+...+22-12
B=12-22+32-42+...-20082-20092
C=3.(22+1)(24+1)(28+1)(216+1)-232
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha
Giải:
a) Sửa đề: 1272 + 146.127 + 732
\(127^2+146.127+73^2=\left(127+7\right)^2=200^2=40000\)
b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^4-1\right)^2=18^8-18^8-1=-1\)
c) \(20^2+18^2+16^2+...+4^2+2^2-\left(19^2+17^2+...+3^2+1\right)\)
\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-...-3^2-1\)
\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+\left(16^2-15^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1\right)\)
\(=20+19+18+17+16+15+...+4+3+2+1\)
\(=\dfrac{\left(20+1\right).20}{2}=210\)
Chúc bạn học tốt!
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=199+195+....+3\)
\(=\frac{\left(199+3\right)\left[\left(199-3\right):4+1\right]}{2}\)
\(=5050\)
\(45^2+40^2-10^2+80.45\)
\(=\left(45+40\right)^2-10^2\)
\(=\left(45+40-10\right)\left(45+40+10\right)\)
\(=75.95=7125\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=\left(2^{32}-1\right)\)
đặt biểu thức \(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\) là A
Ta có:\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^{16}-1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=4^{32}-1\)
\(\Rightarrow A=\dfrac{4^{32}-1}{15}\)
Vậy giá trị biểu thức trên là \(\dfrac{4^{32}-1}{15}\)
\(b,40^2-39^2+38^2-37^2+...+2^2-1^2\)
\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1+2+...+38+39+40\)
\(=\dfrac{\left(40+1\right).40}{2}=\dfrac{41.40}{2}=820\)