phân tích đa thức thành nhân tử:A=(x-1)(x+2)(x-3)(x+4)-144
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SL
2
19 tháng 8 2021
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(t=x^2+5x+4\)
(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)
Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)
19 tháng 8 2021
a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
I
1
AD
1
TA
21 tháng 12 2020
a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)
b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2
7 tháng 10 2021
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
KS
7 tháng 10 2021
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
26 tháng 8 2021
`b)x^3+y^3+z^3-3xyz`
`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`
`=(x+y)^3+z^3-3xy(x+y+z)`
`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`
`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
SO
2
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2021
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
16 tháng 7 2021
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
T
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(x\left(x-1\right)+\left(1-x\right)^2\)
\(=x\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x+x-1\right)\)
\(=\left(x-1\right)\left(2x-1\right)\)
b) \(\left(x+1\right)^2-3\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)-3\right]\)
\(=\left(x+1\right)\left(x+1-3\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
c) \(2x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
22 tháng 8 2021
a: Ta có: \(x^5-x^3+x^2-1\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
c: Ta có: \(16x^4y^2+2xy^5\)
\(=2xy^2\left(8x^3+y^3\right)\)
\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d: Ta có: \(a^3-8+6a^2-12a\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+8a+4\right)\)
e: Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
22 tháng 7 2023
`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!
Trả lời:
A=(x-1)(x+2)(x-3)(x+4)-144
A= (x2-5x-14)(x2-5x-24)-144 (1)
đặt m=x2-5x-14
=> A= m.(m-10)-144
A=m2-10m-144
A= (m-18)(m+8)
thay m vào, ta có:
A= (x2-5x-32)(x2-5x-6)
A=(x2-5x-32)(x+1)(x-6)