a) 101² = (100+1)^2 = 100^2 + 200 + 1 = 10 000 + 200 +1 = 10 201

 b) 199² = (200 -1)^2 = 200^2 - 400 +1 = 40 000 - 400 +1 = 39 601 

; c) 47.53 = (50-3) (50+3) = 50^2 - 3^2= 2500 -9 =2491

a) \(\left(100+1\right)^2=100^2+2\times100\times1+1^2=10000+200+1=10201\)

b)\(=\left(200-1\right)^2=200^2-2\times200\times1+1^2=40000-200+1=39801\)

c)\(=\left(50-3\right)\times\left(50+3\right)=50^2-3^2=2491\)

\(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10000+200+1=10201\)

\(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)

\(47.53=\left(50-3\right).\left(50+3\right)=50^2-3^2=2500-9=2491\)

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,=10201

b,=39601

c,=2491

Nen a<b>c

a,101^2=101.101

             =101.(100+1)

             =(101.100)+(101.1)

             =10100+101

             =10201

b,199^2=199.199

             =199.(200-1)

             =(199.200)-(199.1)

             =39800-199

             =39601

\(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10201\)

\(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=39601\)

\(47.53=\left(50-3\right).\left(50+3\right)=50^2-3^2=2491\)

cảm ơn bạn

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

giải

(47.53)=(47+3)(53-3)

=50.50=2500

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 : 

\(B=\frac{5}{3}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

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 )