PTĐTTNT:
x3 + xy2 - 9xz2 - 2xyz
phân tích đa thức thành nhân tử bằng cách nhóm hạng tử
3) x2 (x+2y) - x - 2y
4) x3 - 4x2 - 9x + 36
5) x2y + xy2 + x2z + y2z + 2xyz
3) \(x^2\left(x+2y\right)-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2y\right)\)
4) \(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)
\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x+3\right)\left(x-3\right)\)
\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
phân tích đa thức thành nhân tử
a)70a+84b-20ab-24b2
b) x2y+xy2+x2z+xz2+y2z+yz2+3xyz
c) x2y+xy2+x2z+xz2+y2z+yz2+2xyz
a: \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(7-2b\right)\left(5a+6b\right)\)
b: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2\right)+\left(y^2z+yz^2\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+2xyz+xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2yz\right)+yz\left(y+z+x\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)^2+yz\left(y+z+x\right)\)
\(=\left(y+z\right)\cdot x\left(x+y+z\right)+yz\left(y+z+x\right)\)
\(=\left(y+z+x\right)\cdot\left(xy+xz+yz\right)\)
c: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2+2xyz\right)+\left(y^2z+yz^2\right)\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2xz\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)
\(=\left(y+z\right)\left(x^2+yz+xy+xz\right)\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
phân tích đa thức thành nhân tử
a)70a+84b-20ab-24b2
b) x2y+xy2+x2z+xz2+y2z+yz2+3xyz
c) x2y+xy2+x2z+xz2+y2z+yz2+2xyz
a) \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(5a+6b\right)\left(7-2b\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)
\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)
\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)
\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)
\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
a, 70a + 84b - 20ab - 24b2
= 14.(5a + 6b) - 4b(5a + 6b)
= (5a + 6b).(14 - 4b)
a, 70a + 84b - 20ab - 24b2
= (70a + 84b) - (20ab + 24b2)
= 14.(5a + 6b) - 4b.(5a + 6b)
= (5a + 6b).(14 - 4b)
Tính tổng của đa thức
P = x2y + x3 – xy2 + 3 và Q = x3 + xy2 – xy – 6.
P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= (x3 + x3) + x2y + (xy2 – xy2) – xy + (3 – 6)
= 2x3 + x2y – xy – 3
Vậy P + Q = 2x3 + x2y – xy – 3.
1, Bậc của đơn thức 3xy2z2 là ?
2, Bậc của đa thức xy2 + 2xyz - 2/3 x5 - 3 ?
1. đơn thức nên bậc 5
2. đa thức nên bậc 5
Phân tích thành nhân tử
1)(2x+y)2-4y2
2)64-125x3
3)x3-2x2y+xy2-4x
4)x3-y3+x2y-xy2
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
Tính tổng và hiệu của hai đa thức P = x2y + x3 – xy2 + 3 và Q = x3 + xy2 – xy – 6
Ta có:
• P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= x2y + (x3 + x3) + (xy2 – xy2) – xy + (3 – 6)
= x2y + 2x3 – xy – 3.
• P – Q = (x2y + x3 – xy2 + 3) – (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 – x3 – xy2 + xy + 6
= x2y + (x3 – x3) – (xy2 + xy2) + xy + (6 + 3)
= x2y – 2xy2 + xy + 9.
Vậy P + Q = x2y + 2x3 – xy – 3; P – Q = x2y – 2xy2 + xy + 9.
\(\text{ P + Q = (x^2y + x^3 – xy^2 + 3) + (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 + x^3 + xy^2 – xy – 6}\)
\(\text{= x^2y + (x^3 + x^3) + (xy^2 – xy^2) – xy + (3 – 6)}\)
\(\text{= x^2y + 2x^3 – xy – 3}\)
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\(\text{P – Q = (x^2y + x^3 – xy^2 + 3) – (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 – x^3 – xy^2 + xy + 6}\)
\(\text{= x^2y + (x^3 – x^3) – (xy^2 + xy^2) + xy + (6 + 3)}\)
\(\text{= x^2y – 2xy^2 + xy + 9}\)
Chứng minh:
(x3+x2y+xy2+y3)(x-y)=x3-y3
(x3+x2y+xy2+y3)(x-y)
=x(x3+x2y+xy2+y3)-y(x3+x2y+xy2+y3)
=x4+x3y+x2y2+xy3-x3y-x2y2+xy3+y4
= x4+y4
đề sai bạn xem lại đề
1. Tính tổng của hai đa thức trong mỗi trường hợp sau :
a, P= x2y + x3 - xy2 +3 và Q= x3 + xy2 - xy - 6
b, M= x2y + 0,5xy3 - 7,5 x3y2 + x3 và N= 3xy3 - x2y + 5,5x3y2
a/ \(P+Q=\left(x^2y+x^3-xy^2+3\right)+\left(x^3+xy^2-xy-6\right)\)
\(=x^2y+x^3-xy^2+3+x^3+xy^2-xy-6\)
\(=\left(x^3+x^3\right)+\left(xy^2-xy^2\right)+\left(3-6\right)+x^2y-xy\)
\(=2x^3+x^2y-xy-3\)
b/ \(M+N=\left(x^2y+0,5xy^3-7,5x^3y^2+x^3\right)+\)
\(\left(3xy^3-x^2y+5,5x^3y^2\right)\)
\(=x^2y+0,5xy^3-7,5x^3y^2+x^3+3xy^3-x^2y+5,5x^3y^2\)
\(=\left(x^2y-x^2y\right)+\left(0,5xy^3+3xy^3\right)+\left(5,5x^3y^2-7,5x^3y^2\right)+x^3\)
\(=3,5xy^3-2x^3y^2+x^3\)