Phân tích các đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp. a) 7x - 6x2-2
b)2x3+3x-5
c) 16x - 5x2-3
d) x3-x+3x2y+3xy2+y3-y
e) 5x2-10xy+5y2-20z2
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15 tháng 10 2021
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
15 tháng 10 2021
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
21 tháng 7 2017
a) Ta có : a2x + a2y - 7x - 7y
= a2(x + y) - (7x + 7y)
= a2(x + y) - 7(x + y)
= (x + y)(a2 - 7)
b) Ta có : x3 + y(1 - 3x2) + x(3x2 - 1) - y3
= x3 - y(3x2 - 1) + x(3x2 - 1) - y3
= x3 - y3 + [x(3x2 - 1) - y(3x2 - 1)]
= x3 - y3 - (3x2 - 1)(x - y)
= (x - y)(x2 + xy + y2) - (3x2 - 1)(x - y)
= (x - y)[(x2 + xy + y2) - (3x2 - 1)]
= (x - y)(x2 + xy + y2 - 3x2 + 1)
= (x - y)(-2x2 + xy + y2 + 1)
21 tháng 7 2017
bài 2:a. \(5x.\left(y^2-2yz+z^2\right)\)
\(=5x.\left(y-z\right)^2\) .......k bít dc chưa
b.\(\left(x^2y-x\right)+\left(xy^2-y\right)\)
\(=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(=\left(xy-1\right).\left(x+y\right)\)
31 tháng 10 2021
1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
31 tháng 10 2021
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
23 tháng 10 2021
11: \(2x^2-12xy+18y^2\)
\(=2\left(x^2-6xy+9y^2\right)\)
\(=2\left(x-3y\right)^2\)
12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)
\(=\left(x^2+x+2\right)\left(x^2+x+1\right)\)
PT
1
CM
30 tháng 6 2019
5 x 2 - 10 x y + 5 y 2 - 20 z 2 = 5 x 2 – 2 x y + y 2 – 4 z 2 = 5 x – y 2 – 2 z 2 = 5 x – y + 2 z x – y – 2 z
21 tháng 10 2021
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
16 tháng 7 2019
h)Ta có : \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt\(x^2+7x+11=y\)
\(=>p\left(x\right)=\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)
Thay \(y=x^2+7x+11\) vào ta có : \(p\left(x\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
16 tháng 7 2019
\(f)m\left(x\right)=x^6+27=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
e)\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=\left(x^2+x\right)\left(x^2+x-2\right)+6\left(x^2+x-12\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)=\left(x^2+x+6\right)\left(x^2-x+2x-2\right)=\left(x^2+x+6\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu a trước đi ạ ^^
a) 7x - 6x2 - 2
= - 6x2 + 7x - 2
= (- 6x2 + 3x) + (4x - 2)
= 3x (- 2x + 1) + 2 (2x-1)
= - 3x ( 2x -1) + 2 (2x - 1)
= ( 2x -1 ) ( - 3x +2 )