\(\text{a) }301^2\\ =\left(300+1\right)^2\\ =300^2+2\cdot300\cdot1+1^2\\ =90000+600+1\\ =90601\\ \)

\(\text{b) }499^2\\ =\left(500-1\right)^2\\ =500^2-2\cdot500\cdot1+1^2\\ =250000-1000+1\\ =249001\\ \)

\(\text{c) }68\cdot72\\ =\left(70-2\right)\left(70+2\right)\\ =70^2-2^2\\=4900-4\\ =4896\\ \)

a) \(301^2\)

\(=\left(300+1\right)^2\)

\(=300^2+600+1\)

\(=90601\)

b) \(499^2\)

\(=\left(500-1\right)^2\)

\(=500^2-1000+1\)

\(=249001\)

c) \(68\cdot72\)

\(=\left(70-2\right)\left(70+2\right)\)

\(=70^2-2^2\)

\(=4896\)

a) \(301^2\)

\(=\left(300+1\right)^2\)

\(=300^2+2.300.1+1^2\)

\(=90000+600+1\)

\(=90601\)

b) \(499^2\)

\(=\left(500-1\right)^2\)

\(=500^2-2.500.1+1^2\)

\(=250000-1000+1\)

\(=249001\)

c) \(68.72\)

\(=\left(70-2\right)\left(70+2\right)\)

\(=70^2-2^2\)

\(=4900-4\)

\(=4896\)

Chúc bạn học tốt!!!

a, \(\left(301\right)^2=\left(300+1\right)^2=300^2+2.300.1+1^2\)

=90000+6000+1=90601

b,\(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2\)

=10000-10000+1=1

a) 301² = ( 300 + 1 )² = 300² + 2.300.1 + 1² = 90601

b) 499² = ( 500 - 1 )² = 500² - 2.500.1 + 1² = 249001

c) 68.72 = ( 70 - 2). ( 70 + 2) = 70² - 4² = 4900 - 16 = 4884

a) 90601

b)249001

c)4896

a=1

b=400

c=-500

a) - 287+ 499+ (- 499)+ 285 

= ( - 287 + 285 ) + [ 499 + ( - 499 ) ]

= -2 + 0 = -2

b) ( 326- 43) + ( 174- 57)

= 326 - 43 + 174 - 57

= ( 326 + 174 ) - ( 43 + 57 )

=  500 - 100

= 400

c) ( 351- 875) - ( 125 - 149)

= 351 - 875 - 125 + 149

= ( 351 + 149 ) - ( 875 + 125)

= 500 - 1000

=  - 500

d) (-8) . 25 .(-2) .4 (-5) .125

= [ ( - 8 ) . 125 ] . (25 . 4) . [( - 2) . ( - 5)

= 1000 .100 . 10

= 1000000

e) (-167). 83+ 167. (-17) -33

= 167 . [ ( - 83 ) + ( - 17) ] - 33

= 167 . ( - 100) -33

= - 16700 -33

= - 16733

g) (- 284). 172+ (-284). (-72)

= ( - 284 ) . ( 172 - 72)

= - 284 . 100

= - 28400

~ Học tốt

# Chiyuki Fujito

Bài1:

\(\left(3+xy^2\right)^2=81+6xy^2+x^2y^4\)

Các câu sau tương tự

Bài2:

\(a,\left(4x^2+4xy+y^2\right)\)

=\(\left(2x+y\right)^2\)

b)\(9m^2+n^2-6mn=\left(3m-n\right)^2\)

c)\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)

d)\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

Bài3:

\(a,301^2=\left(300+1\right)^2=900+600+1=1501\)

b/\(499^2=\left(500-1\right)^2=2500-1000+1=1501\)

c/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)

Bài 1:

a, \(\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)

b, \(\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)

c, \(\left(a-b\right)^2.\left(a+b\right)^2=\left[\left(a-b\right)\left(a+b\right)\right]^2\)

\(=\left(a^2-b^2\right)^2=a^4-2a^2b^2+b^4\)

Chúc bạn học tôt!!!

62 . 58 = (60 + 2)(60 - 2) = 60\(^2\) - 2\(^2\) = 3600 - 4 = 3596

199\(^2\) = (200 -1)\(^2\) = 200\(^2\) - 2.200.1 + 1\(^2\) = 40 000 - 400 + 1 = 39601

499\(^2\) = (500 - 1)\(^2\) = 500\(^2\) - 2.500.1 + 1\(^2\) = 250 000 - 1000 + 1 = 249 001

299 . 301 = (300 - 1)(300 + 1) = 300\(^2\) - 1\(^2\) = 90 000 - 1 = 89 999

Học tốt

Đúng thì k cho mk nhé

Trả lời:

+, \(62.58=\left(60+2\right)\left(60-2\right)=60^2-2^2=3600-4=3596\)

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

+, \(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2=250000-1000+1=249001\)

+, \(299.301=\left(300-1\right)\left(300+1\right)=300^2-1=90000-1=89999\)

62 * 58  =   ( 60 + 2 ) * ( 60 - 2 ) = 60^2  -  2^2  =  3600 - 4 = 3596

199^2   =  ( 200 - 1 )^2 = 200^2  - 2 * 200 * 1 + 1^2 =  40000 - 400 + 1 = 39601

499^2  =  (  500  - 1 )^2  =  500^2 - 2* 500 *1  + 1^2  =  250000 - 1000 + 1 = 249001

299 * 301=  (  300 - 1 ) * ( 300 + 1 ) = 300^2 - 1^2  = 90000 - 1 = 89999

a,=15050
b, = 10000

a) 50 . 301

= 50 .(300 + 1)

= 50 . 300 + 50

= 50 . 3 . 100 + 50

= 150 . 100 + 50

= 15000 + 50

= 15050

b) 32 . 48 + 52 . 32 + 68 . 100

= 32 . (48 + 52) + 68 . 100

= 32 . 100 + 68 . 100

= 100 . (32 + 68)

= 100 . 100

= 10 000

 a) 1 + 2 + 3 + ..... + 2006 + 2007

= ( 2007 + 1 ) . 2007 : 2 = 2015028

b) 132 + 128 + 124 +72 + 68

= ( 132 + 68 ) + ( 128 + 72 ) + 124

= 200 + 200 + 124

= 524

a, 1+2+3+4+....+2001+2002=(2002+1).[(2002-1)+1] :2

                                        =2003.1001

                                       = 2005003

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 1: 

c) Ta có: \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Leftrightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Leftrightarrow3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow2B=1-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)