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1/ a, \(=4^2-a^2=\left(4-a\right)\left(4+a\right)\)
b, \(=\left(a+b\right)^2-\left(2c\right)^2=\left(a+b-2c\right)\left(a+b+2c\right)\)
2/ a, \(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10000+200+1=10201\)
b, \(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)
c, \(47.53=\left(50-3\right)\left(50+3\right)=50^2-3^2=2500-9=2491\)
a)Ta có \(101^2\)=\(\left(100+1\right)^2\)=10000+200+1
=10201
b)\(199^2\)=\(\left(200-1\right)^2=40000-400+1\)=39601
c)47.53=\(\left(50-3\right)\left(50+3\right)=50^2-3^2\)=2500-9=2491
Bài làm:
a) 101² Sử dụng hằng đẳng thức thứ 1 ta có:
= (100 + 1)²
= 100² + 2.100.1 + 1²
= 10000 + 200 + 1
= 10201
b) 199² Sử dụng hằng đẳng thức thứ 2 ta có:
= (200 - 1)²
= 200² - 2.200.1 + 1²
= 40000 - 400 + 1
= 39601
c) 47 . 53 Sử dụng hằng đẳng thức thứ 3 ta có:
= (50 - 3)(50 + 3)
= 50² - 3²
= 2500 - 9
= 2491.
a) 1012=(100+1)2=1002+2.50.2+12=10000+200+1=10201
b)1992=(200-1)2=2002 -2.200.1+12=40000-400+1=39601
c) 47.53=(50-3)(50+3)=502-32=2500-9=2491
a) 1012 =(100+1)2 =10000+1=10001
b) 1992 =(199+1)2 =2002 =40000
c) 47.53=(40+7 .50+3)=20000+10=20010
cau 1 \(x^2+6xy+9y^2=\left(x+3y\right)^2\)( binh phuong cua mot tong)
\(x^2-10xy+25y^2=\left(x-5y\right)^2\)( binh phuong cua mot hieu )
\(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10201\)
\(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=39601\)
\(47.53=\left(50-3\right).\left(50+3\right)=50^2-3^2=2491\)
d )
\(B=5\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{64}-1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{128}-1\right)\)
Sửa lại dấu \(\Rightarrow\)dòng 3 :
\(B=\frac{5}{3}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)