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A = 76.56-(353+2)(353-2)
= 356- 356-2
= -2
B = 19972-1996.1998
=19972-( 1997-1)(1997+1)
= 19972 - 1997-1
ĐẾN đey chịu ^^"
a) \(49.51=\left(50-1\right)\left(50+1\right)=50^2-1^2=2500-1=2499\)
b) \(29.31=\left(30-1\right)\left(30+1\right)=30^2-1^2=900-1=899\)
c) \(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10000+200+1=10201\)
d) \(99^2+2.99+1=\left(99+1\right)^2=100^2=10000\)
e) \(\left(10^2+8^2+6^2+4^2+2^2\right)-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(=10^2-9^2+8^2-7^2+6^2-5^2+4^2-3^2+2^2-1^2\)
\(=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+\left(6-5\right)\left(6+5\right)+\)
\(\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(=10+9+8+7+6+5+4+3+2+1=55\)
f) \(1998^2-1997.\left(1998+1\right)=1998^2-\left(1998-1\right)\left(1998+1\right)\)
\(=1998^2-1998^2+1=1\)
A=6+52+53+54+...+51996+51997
A = 1 + 5 + 52 + 53 + ... + 51996 + 51997
5A = 5 + 52 + 53 + 54 + ... + 51997 + 51998
5A - A = ( 5 + 52 + 53 + 54 + ... + 51997 + 51998 ) - ( 1 + 5 + 52 + 53 + ... + 51996 + 51997 )
4A = 51998 - 1
\(\Rightarrow A=\frac{5^{1998}-1}{4}\)
\(A=6+5^2+5^3+...+5^{1996}+5^{1997}\\ A=1+5+5^2+5^3+...+5^{1996}+5^{1997}\)
\(5A=5+5^2+5^3+...+5^{1996}+5^{1998}\)
\(5A-A=\left(5+5^2+5^3+...+5^{1996}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)
\(4A=5^{1998}-1\\ A=\frac{5^{1998}-1}{4}\)
a) \(37^2+2\cdot37\cdot13+13^2\)
\(=1369+962+169\)
\(=2500\)
b) \(35^2+24^2-48\cdot35\)
\(=125+576-1680\)
\(=121\)
c) sai quy luật
Ta có: \(A=6+5^2+5^3+5^4+...+5^{1996}+5^{1997}=1+5+5^2+5^3+...+1^{1997}\)
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\)
\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)
\(\Rightarrow4A=5^{1998}-1\Rightarrow A=\dfrac{5^{1998}-1}{4}\)
Vậy ...
\(A=165^2+70.165+35^2=165^2+2.35.165+35^2=\left(165+35\right)^2=200^2=40000\)
\(C=5^4.7^4-\left(35^2-1\right)\left(35^2+1\right)=35^4-[\left(35^2\right)^2-1^2]=35^4-35^4+1=1\)
Bài 2:
a: \(\left(a-b-2\right)^2-\left(2a-2b\right)\left(a-b-2\right)+a^2-2ab+b^2\)
\(=\left(a-b\right)^2-4\left(a-b\right)+4+\left(a-b\right)^2-2\left(a-b\right)\left(a-b-2\right)\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left[\left(a-b\right)^2-2\left(a-b\right)\right]\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left(a-b\right)^2+4\left(a-b\right)\)
\(=4\)
b: \(\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(=2^{512}-1+1=2^{512}\)
c: \(24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}\)
=-1
Ta có: A = 6 + 52 + 53 + 54 + ... + 51996 + 51997
A = 1 + 5 + 52 + 53 + ... + 51996 + 51997
5A = 5(1 + 5 + 52 + 53 + ... + 51996 + 51997)
5A = 5 + 52 + 53 + 54 + ... + 51997 + 51998
5A - A = (5 + 52 + 53 + 54 + ... + 51997 + 51998) - (1 + 5 + 52 + 53 + ... + 51996 + 51997)
4A = 51998 - 1
A = \(\frac{5^{1998}-1}{4}\)
A= 6 + 52+ 53+ 54 + ..... + 5 1996+ 51997
=>5A=5+52+53+54+...+51997+51998
=5A-A=(5+52+53+54+...51997+51998)-(1+5+52+53+...+51996+51997)
=4A=51998-1=>A=\(\frac{5^{1998}-1}{4}\)
Vậy ...
hc tốt