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a) 262 + 52.24 + 242
= 262 + 2.26.24 + 242
= ( 26 + 24 )2
= 502 = 2500
b) 30032 - 32
= ( 3003 + 3 ) ( 3003 - 3 )
= 3006 . 3000 = 9018000
c) 872 + 732 - 272 -132
= ( 872 - 132 ) + ( 732 - 272 )
= [ ( 87 + 13 )( 87- 13 )] + [ ( 73 - 27 )( 73 + 27 ) ]
= ( 100 . 74 ) + ( 46 . 100 )
= 7400 + 4600 = 12000
d)792 - 79.58 + 292
= 792 - 2.79.29 + 292
= ( 79 - 29 )2
= 502 = 2500
a) 262 + 52 . 24 + 242 = 262 + 2 . 26 . 24 + 242
= ( 26 + 24 )2
= 502
= 2500
b) 30032 - 32 = ( 3003 - 3 ) . ( 3003 + 3 )
= 3000. 3006
= 9018000
c) 872 + 732 - 272 - 132 = ( 872 - 272 ) + ( 732 - 132 )
= ( 87 - 27 ) . ( 87 + 27 ) + ( 73 - 13 ) . ( 73+13)
= 60 . 114 + 60 . 86
= 60 . ( 114 + 86 )
= 60 . 200
= 12000
d) 792 - 79 . 58 + 292 = 792 - 2 . 79 . 29 + 292
= ( 79 - 29 )2
= 502
= 2500
a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)\)
= 400
b) \(87^2+73^2-27^2-13^2\)
\(\Leftrightarrow\left(87^2-13^2\right)\)+\(\left(73^2-27^2\right)\)
\(\Leftrightarrow\left(87+13\right)\left(87-13\right)+\left(73+27\right)\left(73-27\right)\)
\(\Leftrightarrow7400+4600=12000\)
a) Ta có: \(37^2+2\cdot37\cdot13+13^2\)
\(=\left(37+13\right)^2=50^2=2500\)
b) Ta có: \(201^2=\left(200+1\right)^2\)
\(=200^2+2\cdot200+1\)
\(=40000+200+1=40201\)
c) Ta có: \(37\cdot43=\left(40+3\right)\cdot\left(40-3\right)\)
\(=40^2-3^2=1600-9=1591\)
Giải:
1). \(25^2-15^2\)
\(=\left(25-15\right)\left(25+15\right)\)
\(=10.40\)
\(=400\)
2). \(87^2+73^2-27^2-13^2\)
\(=\left(87^2-13^2\right)+\left(73^2-27^2\right)\)
\(=\left(87-13\right)\left(87+13\right)+\left(73-27\right)\left(73+27\right)\)
\(=74.100+46.100\)
\(=100\left(74+46\right)\)
\(=100.120\)
\(=12000\)
Chúc bạn học tốt!!!
a) 732 - 272
= ( 73 + 27 ) ( 73 - 27 )
= 100 . 46
= 4600
b) 372 - 132
= ( 37 + 13 ) ( 37 - 13 )
= 50 . 24
= 1200
c) 20022 - 22
= ( 2002 + 2 ) ( 2002 - 2 )
= 2004 . 2000
= 4008000
câu c bạn viết thiếu mjk viết lại
a)73^2 - 27^2
=(73+27)(73-27)
=100.46
=4600
b)37^2 - 13^2
=(37+13)(37-13)
=50.24
=1200
c)2002^2 - 2^2
=(2002+2)(2002-2)
=2004.2000
=4008000
bài 1
a(x+y)2-(x-y)2
=[(x+y)-(x-y)][(x+y)+(x-y)]
=(x+y-x+y)(x+y+x-y)
=2y.2x
b,(3x+1)2-(x+1)2
=[(3x+1)-(x+1)][(3x+1)+(x+1)]
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
4x.(x+10
bài 2
x3-0,25x=0
=>x(x2-0,25)=0
=>x=0 hoặc x2-0,25=0
=> x=0 hoặc x=\(\pm0,5\)
\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)
Vậy phương trình có nghiệm \(x=2004\)
\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)
Bài giải:
a) 732 – 272 = (73 + 27)(73 – 27) = 100 . 46 = 4600
b) 372 - 132 = (37 + 13)(37 – 13) = 50 . 25 = 100 . 12 = 1200
c) 20022 – 22 = (2002 + 2)(2002 – 2) = 2004 . 2000 = 400800
câu b tại sao lại suy ra đc 100.12 vậy chị ?