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\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
\(a,36-4a^2+20ab-25b^2\)
\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,5a^2-10a^2b+5ab^2-10a+10b\)
\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)
\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)
\(=\left(a-b\right)\left(5a-5ab-10\right)\)
c) x2 + 2xy + y2 – xz – yz = (x + y)2 – z(x + y) = (x + y)(x + y – z)
c) x2 + y2 + xz + yz + 2xy
= (x2 + 2xy + y2) + (xz + yz)
= (x + y)2 + z(x + y)
= (x + y)(x + y + z)
d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)
a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)
\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b2 thành 27b3 vì mình nghĩ nhầm đề)
\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)
\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
c)\(x^3+3x^2+3x+1-3x^2-3x=\left(x+1\right)^3-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
a) \(3xy^2-12xy+12x\)
\(=3x\left(y-4y+4\right)\)
b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)
\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)
\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)
\(=3xy\left[\left(x-1\right)^2-\left(y+a\right)^2\right]\)
\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)
c) \(36-4a^2+20ab-25b^2\)
\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
d) \(5a^3-10a^2b+5ab^2-10a+10b\)
\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)
\(=5a\left(a-b\right)^2-10\left(a-b\right)\)
\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)
\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)
\(=5\left(a-b\right)\left(a^2-ab-2\right)\)
a. 3xy2-12xy+12x
= 3x(y2-4y+4)
= 3x(y-2)2
b. 3x3y-6x2y-3xy3-6axy2-3a2xy+3xy
= 3xy( x2-2x-y2-2ay-a2+1)
= 3xy ((x2-2x+1)-(a2-2ay-y2))
=3xy((x-1)2-(a-y)2)
= 3xy((x-1+a-y)(x-1-(a-y))
=3xy(x-1+a-y)(x-1-a+y)
d. =( 5a(a2-2ab+b2))-(10(a+b))
= 5a(a-b)2-10(a-b)
=5a(a-b)(a-b)-10(a-b)
=(a-b)(5a(a-b)-10)
Hình như mik làm sai hết rồi
a) Cách 1.
Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)
= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).
Cách 2.
Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)
= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).
b) Biến đổi được a 4 - 9 rt 3 + a 2 -9a = (a- 9)a( a 2 +1).
c) Biến đổi được 3 x 2 + 5y - 3xy + (-5x) = (x - y)(3x - 5).
d) Biến đổi được x 2 - (a + b)x + ab = (x- a)(x - b).
e) Ta có 4 x 2 - 4xy + y 2 – 9 t 2 = ( 2 x - y ) 2 - ( 3 t ) 2
= (2x - y - 3t )(2x - y + 31).
g) Ta có x 3 - 3 x 2 y + 3 xy 2 - y 3 - z 3
= ( x - y ) 3 - z 3 = (x - y - z)( x 2 + y 2 + z 2 - 2xy + xz - yz).
h) Ta có x 2 - y 2 + 8x + 6y+ 7 = ( x 2 +8x + 16) - ( y 2 - 6y+ 9)
= ( x + 4 ) 2 - ( y - 3 ) 2 =(x-y + 7)(x + y + l).
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)