tính nhanh
a) 301^2
b)499^2
c) 68 . 72
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\(\text{a) }301^2\\ =\left(300+1\right)^2\\ =300^2+2\cdot300\cdot1+1^2\\ =90000+600+1\\ =90601\\ \)
\(\text{b) }499^2\\ =\left(500-1\right)^2\\ =500^2-2\cdot500\cdot1+1^2\\ =250000-1000+1\\ =249001\\ \)
\(\text{c) }68\cdot72\\ =\left(70-2\right)\left(70+2\right)\\ =70^2-2^2\\=4900-4\\ =4896\\ \)
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, \(\left(301\right)^2=\left(300+1\right)^2=300^2+2.300.1+1^2\)
=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) - 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
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/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
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:
+, \(62.58=\left(60+2\right)\left(60-2\right)=60^2-2^2=3600-4=3596\)
+, \(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)
+, \(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2=250000-1000+1=249001\)
+, \(299.301=\left(300-1\right)\left(300+1\right)=300^2-1=90000-1=89999\)
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+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: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)
\(=75\cdot\left(-4\right)+603\)
\(=603-300=303\)
Giải:
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!!!