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a: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b: \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
\(x^6-2x^3y+y^2=\left(x^3-y\right)^2\)
b: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
\(-a^2-2a-1=-\left(a+1\right)^2\)
mk ghi đáp án, ko phân tích đc thì IB mk
a) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b) \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
c) \(x^6+y^2-2x^3y=\left(x^3-y\right)^2\)
d) \(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(3x^2+y^2\right)\)
e) \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
f) \(-a^2-2a-1=-\left(a+1\right)^2\)
g) \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)
h) \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)
i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)
a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
\(=\left[\left(x+a\right)\left(x+4a\right)\right]\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax+5a^2\right)^2-\left(a^2\right)^2+a^4\)
\(=\left(x^2+5ax+5a^2\right)^2\)
b) \(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)^2+2\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)
a) ta có : \(x^2+6xy+9y^2=x^2+2.x.3y+\left(3y\right)^2=\left(x+3y\right)^2\)
b) ta có : \(4a^4-4a^2b^2+b^4=\left(2a^2\right)^2-2.2a^2.b^2+\left(b^2\right)^2=\left(2a^2-b^2\right)^2\)
c) ta có : \(x^6+y^2-2x^3y=\left(x^3\right)^2-2.x^3.y+y^2=\left(x^3-y^2\right)^2\)
d) ta có : \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right)\)
\(=2y\left(x^2+y^2+2xy+2x^2-2y^2+x^2+y^2-2xy\right)\)
\(=2y\left(4x^2\right)=8x^2y\)
e) ta có : \(25x^4-10x^2y^2+y^4=\left(5x^2\right)^2-2.5x^2.y^2+\left(y^2\right)^2=\left(5x^2-y^2\right)^2\)
f) ta có : \(-a^2-2a-1=-\left(a^2+2a+1\right)=-\left(a+1\right)^2\)
g) ta có : \(27b^3-8a^3=\left(3b\right)^3-\left(2a\right)^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)
i) ta có : \(16x^2-9\left(x+y\right)^2=\left(4x\right)^2-\left(3\left(x+y\right)\right)^2\)
\(=\left(4x-3x-3y\right)\left(4x+3x+3y\right)=\left(x-3y\right)\left(7x+3y\right)\)
Phối hợp cả 3 phương phép để phân tích các đa thức sau thành phân tử:
a) 36 - 4a2 + 20ab - 25b2
= 36 - (4a2 - 20ab + 25b2)
= 62 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
b) a3 + 3a2 + 3a + 1 - 27b3
= (a + 1)3 - (3b)3
= (a + 1 - 3b)[(a + 1)2 + 3b(a + 1) + 9b2]
= (a + 1 - 3b)(a2 + 2a + 1 + 3ab + 3b + 9b2)
c) x2 + 2xy + y2 - xz - yz
= (x + y)2 - z(x + y)
= (x + y)(x + y - z)
d) 5a3 - 10a2b + 5ab2 - 10a + 10b
= 5(a3 - 2a2b + ab2 - 2a + 2b)
= 5[a(a2 - 2ab + b2) - 2(a - b)]
= 5[a(a - b)2 - 2(a - b)]
= 5(a - b)(a2 - ab - 2)
Bài 1:
a: \(3x\left(x-a\right)+4a\left(a-x\right)\)
=3x(x-a)-4a(x-a)
=(x-a)(3x-4a)
b: \(x^2\left(y^2+z\right)+y^3+yz\)
\(=x^2\left(y^2+z\right)+y\left(y^2+z\right)\)
\(=\left(x^2+y\right)\left(y^2+z\right)\)
c: \(3x^2\left(x+1\right)-5x\left(x+1\right)^2+4\left(x+1\right)\)
\(=\left(x+1\right)\left[3x^2-5x\left(x+1\right)+4\right]\)
\(=\left(x+1\right)\left(3x^2-5x^2-5x+4\right)\)
\(=\left(x+1\right)\left(-2x^2-5x+4\right)\)
Phân tích các đa thức sau thành nhân tử ... c) 6x(x+y)^2+3x^2y(x+y). 2: .... x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x. x3 - 9x2 ..... Phân tích các đa thức sau thành nhân tử : a,x^3+5x^2+8x+4 b, x^3-9x^2+6x+16 .
\(a,36-4a^2+20ab-25b^2\)
\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,5a^2-10a^2b+5ab^2-10a+10b\)
\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)
\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)
\(=\left(a-b\right)\left(5a-5ab-10\right)\)
a) = 9x2 - ( y2 - 10y + 25y2 ) = ( 3x )2 - ( y - 5 )2 = ( 3x - y + 5 )( 3x + y - 5 )
b) = ( x3 - 8 ) - ( x2 - 4x + 4 ) = ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )2 = ( x - 2 )( x2 + x + 6 )
c) = ( 4a2 - 4a + 1 ) - ( b2 - 2bc + c2 ) = ( 2a - 1 )2 - ( b - c )2 = ( 2a - b + c - 1 )( 2a + b - c - 1 )
d) = ( a3 + 3a2 + 3a + 1 ) - 27b3 = ( a + 1 )3 - ( 3b )3 = ( a - 3b + 1 )( a2 + 9b2 + 3ab + 3b )
a. \(9x^2-\left(y^2-10y+25\right)=9x^2-\left(y-5\right)^2=\left(3x-y+5\right)\left(3x+y-5\right)\)
b.\(x^3-8-x^2+4x-4=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)^2=\left(x-2\right)\left(x^2+x+6\right)\)
c.\(\left(4a^2-4a+1\right)-\left(b^2-2bc+c^2\right)=\left(2a-1\right)^2-\left(b-c\right)^2=\left(2a-1+b-c\right)\left(2a-1-b+c\right)\)
d.\(\left(a^3+3a^2+3a+1\right)-27b^3=\left(a+1\right)^3-\left(3b\right)^3=\left(a+1-3b\right)\left[\left(a+1\right)^2+3b\left(a+1\right)+9b^2\right]\)