Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
A = x2(x - 1) + 6(1 - x)
A = x3 - x2 + 6 - 6x
A = (x3 - 6x) - (x2 - 6)
A = x.(x2 - 6) - (x2 - 6)
A = (x - 1)(x2 - 6)
C = x2 + 2xy + y2 - yz - xz
C = (x + y)2 - z.(x + y)
C = (x + y - z).(x + y)
a) xy – 3x + 2y – 6
= (xy - 3x) + (2y - 6)
= x(y - 3) + 2(y - 3)
= (y - 3)(x + 2)
b) x2y + 4xy + 4y – y3
= y(x2 + 4x + 4 - y2)
= y[(x2 + 4x + 4) - y2]
= y[(x + 2)2 - y2]
= y(x + 2 + y)(x + 2 - y)
c) x2 + y2 + xz + yz + 2xy
= (x2 + 2xy + y2) + (xz + yz)
= (x + y)2 + z(x + y)
= (x + y)(x + y + z)
d) x3 + 3x2 – 3x – 1
= (x3 - 1) + (3x2 - 3x)
= (x - 1)(x2 + x + z) + 3x(x - 1)
= (x - 1)(x2 + 4x + 1)
a )
\(xy-3x+2y-6\)
\(=\left(xy+2y\right)-3x-6\)
\(=y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(y-3\right)\left(x+2\right)\)
b )
\(x^2y+4xy+4y-y^3\)
\(=y\left(x^2+4x+4-y^2\right)\)
\(=y\left[\left(x+2\right)^2-y^2\right]\)
\(=y\left(x+2-y\right)\left(x+2+y\right)\)
c )
\(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(5x^2y^3-25x^2y^2+10x^2y^4=5x^2y^2\left(y-5+2y^2\right)\)
\(12a^4-24a^2b^2-6ab=6a\left(2a^3-4ab^2-3b\right)\)
mk chỉnh đề
\(-25x^6-y^8+10x^3y^4=-\left(5x^3-y^4\right)^2\)
\(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
a) \(2x-6y=2\left(x-3y\right)\)
b) \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
c) \(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
d) \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e) \(x^3-10x^2+25x=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\)
f) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
a)
\(-25x^6-y^8+10x^3y^4=-(25x^6+y^8-10x^3y^4)\)
\(=-[(5x^3)^2-2.5x^3y^4+(y^4)^2]\)
\(=-(5x^3-y^4)^2\)
b) \(49(y-4)^2-9(y+2)^2=7^2(y-4)^2-3^2(y+2)^2\)
\(=(7y-28)^2-(3y+6)^2=(7y-28-3y-6)(7y-28+3y+6)\)
\(=(4y-34)(10y-22)=4(2y-17)(5y-11)\)
c)
\(x^2+2xy+y^2-xz-yz=(x^2+2xy+y^2)-(xz+yz)\)
\(=(x+y)^2-z(x+y)=(x+y)(x+y-z)\)
d)
\(1+(a+b+c)+(ab+bc+ac)+abc\)
\(=(1+a+b+ab)+(bc+ac+abc+c)\)
\(=(1+a+b+ab)+c(b+a+ab+1)\)
\(=(a+b+ab+1)(c+1)\)
\(=[(a+ab)+(b+1)](c+1)\)
\(=[a(b+1)+(b+1)](c+1)\)
\(=(a+1)(b+1)(c+1)\)