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( a+ b+ c)^2 = a^2 + b^2 + c^2 + 2ab+ 2bc+ 2ac
a, ( x + y - 2)^2 = x^2 + y^2 + 4 + 2xy - 4x - 4y
bl, ( 2x + 3y+ 5)^2 = 4x^2 + 9y^2 + 25 + 12xy + 20x + 30y
c < ( 3x - y + 2)^2 = 9x^2 + y^2 + 4 -6xy + 12x - 4y
Bài 1.
a) ( 3x + 4y )2 = ( 3x )2 + 2.3x.4y + ( 4y )2 = 9x2 + 24xy + 16y2
b) ( x2 + 1 )2 = ( x2 )2 + 2.x2.1 + 12 = x4 + 2x2 + 1
c) ( 3 - 2y )2 = 32 - 2.3.2y + ( 2y )2 = 9 - 12y + 4y2
d) ( xy2 - 2 )2 = ( xy2 )2 - 2.xy2.2 + 22 = x2y4 - 4xy2 + 4
Bài 2.
a) x2 - 9 = x2 - 32 = ( x - 3 )( x + 3 )
b) 25 - 4y2 = 52 - ( 2y )2 = ( 5 - 2y )( 5 + 2y )
c) 9x4 - 4y2 = ( 3x2 )2 - ( 2y )2 = ( 3x2 - 2y )( 3x2 + 2y )
d) ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )
B1:
a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)
b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)
c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)
d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)
B2:
a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)
c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)
d) \(\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
Bài 1 : Viết các đa thức sau dưới dạng lập phương của một tổng hoặc lập phương của một hiệu
a,8x3+12x2y+6xy2+y38x3+12x2y+6xy2+y3
= (2x)3 + 3.(2x)2.y + 3.2x.y2 + y3
= ( 2x + y )3
b,x3+3x2+3x+1x3+3x2+3x+1
= x3 + 3.x2.1 + 3.x.12 + 13
=(x + 1)3
c, x3−3x2+2x−1x3−3x2+2x−1
= x3 - 3.x2.1+ 3.x.12 - 13
= (x - 1)3
d,27+27y2+9y4+y6
= 33 + 3.32.y2 + 3.3.y4 + (y2)3
= ( 3 + y2 ) 3
(a+b+c)3= (a+b)3+3(a+b)2c+3(a+b)c2+c2
=a3+3a2b+3ab2+b2+3(a+b)c(a+b+c)+c2
=a3+b3+c3+3ab(a+b)+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)[ab+c(a+b+c)]
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[(ab+ac)+(bc+c2)]
=a3+b3+c3+3(a+b)[a(b+c)+c(b+c)]
=a3+b3+c3+3(a+b)(b+c)(c+a)
Vậy (a+b+c)3 = a3 + b3 + c3 + 3(a+b)(b+c)(c+a)
a) \(=a\left(a+b\right)\left(-ab\right)\left(a-b\right)=-a^2b\left(a^2-b^2\right)\)
b) \(=\left(3a-1\right)^2+2\left(3a-1\right)\left(3a+1\right)+\left(3a+1\right)^2=\left(3a-1+3a+1\right)^2=\left(6a\right)^2=36a^2\)
c) \(=\left(a^2+b^2\right)^2-a^2b^2-\left(a^4+b^4\right)=a^4+b^4+2a^2b^2-a^2b^2-a^4-b^4=a^2b^2\)
nhớ LI KE
1. \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)
2. \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
3. \(x^2-1=x^2-1^2=\left(x-1\right)\left(x+1\right)\)
4. \(8+x^3=2^3+x^3=2^3+3.2^2.x+3.2.x^2+x^3\)
\(=8+12x+6x^2+x^3\)
5. \(8x^3+27=\left(2x\right)^3+3^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
6. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
7. \(\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\)
8. \(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc\)