x3 + x2y - x2z - xyz
ptích đa thức thành nhân tử = cách nhóm hạng tử
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 (bằng phương pháp nhóm hạng tử)
a/ x3 - x2y + 7x – 7y
\(x^3-x^2y+7x-7y=\left(x^3-x^2y\right)+\left(7x-7y\right)=x^2\left(x-y\right)+7\left(x-y\right)=\left(x-y\right)\left(x^2+7\right)\)
\(x^3-x^2y+7x-7y\)
\(=x^2\left(x-y\right)+7\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x^2+7\right)\)
Phân tích các đa thức sau thành nhân tử
a,3x2 + 6xy + 3y2 - 3z
b,,x3 + x2y - x2z - xyz đ
`@` `\text {Ans}`
`\downarrow`
`a,`
`3x^2 + 6xy + 3y^2 - 3z`
`= 3*x^2 + 3*2xy + 3y^2 - 3z`
`= 3(x^2 + 2xy + y^2 - z)`
`b,`
`x^3 + x^2y - x^2z - xyz`
`= x(x + y)(x-z)`
Phân tích các đa thức sau thành nhân tử:
a/ x2 – 3x – 4xy + 12y b/ x3 – 4x2 + 4x -1
c/ x – y – ax + ay d/ x2 – 4 + ( x + 2)2
e/x3 + x2y – x2z – xyz f/ x2 – y2 – 2x – 2y
a: \(=x\left(x-3\right)-4y\left(x-3\right)\)
=(x-3)(x-4y)
d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2+x+2\right)\)
=2x(x+2)
\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)
phân tích đa thức thành nhân tử :
a) x2 – y2 + 11x – 11y
b) x3 + x2y + yz2 - xyz + z3
\(a,=\left(x-y\right)\left(x+y\right)+11\left(x-y\right)=\left(x-y\right)\left(x+y+11\right)\\ b,=\left(x+z\right)\left(x^2-xz+z^2\right)+y\left(x^2+z^2-xz\right)\\ =\left(x^2-xz+z^2\right)\left(x+y+z\right)\)
a. x2 - y2 + 11x - 11y
= (x + y)(x - y) + 11(x - y)
= (x + y + 11)(x - y)
b. Mik ko hiểu đề lắm
x2y + xy2 - x- y
phân tích đa thức thành nhân tử theo pp nhóm hạng tử
\(X^2y+xy^2-x-y\)
\(=xy(x+y)-(x+y)=(xy-1)(x+y)\)
\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
\(x^2y+xy^2-x-y=\left(x^2y+xy^2\right)-\left(x+y\right)=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
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)
phân tích đa thức thành nhân tử(2 nhóm hạng tử)
x3 -x2 +7x-7
\(x^3-x^2+7x-7=x^2\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(x^2+7\right)\)
Bài 3: Phân tích các đa thức sau thành nhân tử bằng phương pháp nhóm các hạng tử
a) x4-x3-x+1 b)x2y+xy2-x-y
c)ax2+a2y-7x-7y d)ax2+ay-bx2-by
e)x4+x3+x+1 g)x2-2xy+y2-xz+yz
h)x2-y2-x+y i)x2-4+2x+1
giúp mình với,mình cần gấp mn ơi
a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
Cho biểu thức :
B = x3 + x2z + y2z - xyz + y3
a) hãy phân tích B thành nhân tử
b) chứng minh rằng nếu x+y+z=1 thì B \(\ge\) 0
a) \(B=x^3+x^2z+y^2z-xyz+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x^2-xy+y^2\right)\left(x+y+z\right)\)
b) \(B=\left(x^2-xy+y^2\right)\left(x+y+z\right)=x^2-xy+y^2\)
\(=x^2-2.x.\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)
Dấu bằng xảy ra khi \(x=y=0\)