Phần tự luận (8 điểm)
Phân tích thành nhân tử
a ) x 6 – x 4 + 2 x 3 + 2 x 2 b ) 4 x 4 + y 4
K
Khách
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TV
1
7 tháng 10 2021
a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)
TT
1
26 tháng 8 2021
a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)
b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)
Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)
c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)
Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)
d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)
Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
DH
Phân tích thành nhân tử:
a) 8.x^3+y^3
b) 8.x^3-125.y^3
c) (5.x+4)^2-(2.x+1)^2
d)x^3+6.x^2y+12x.y^2+8.y^3
1
CW
12 tháng 9 2016
a) \(8x^3+y^3\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
b) \(8x^3-125y^3\)
\(=\left(2x-5y\right)\left(4x^2+10xy+25y^2\right)\)
c) \(\left(5x+4\right)^2-\left(2x+1\right)^2\)
\(=\left(3x+3\right)\left(7x+5\right)\)
\(=3\left(x+1\right)\left(7x+5\right)\)
d) \(x^3+6x^2y+12xy^2+8y^3\)
\(=\left(x+2y\right)^3\)
6 tháng 9 2017
Ta có : 5x(x - 2y) + 2(2y - x)2
= 5x(x - 2y) + 2(x - 2y)2 (vì (2y - x)2 = (x - 2y)2 )
= (x - 2y)[5x + 2(x - 2y)]
= (x - 2y)(5x + 2x - 4y)
= (x - 2y)(7x - 4y)
b) 7x(y - 4)2 - (4 - y)3
= 7x(y - 4)2 - (4 - y)2(4 - y)
= 7x(y - 4)2 - (y - 4)2(4 - y)
= (y - 4)2(7x - 4 + y)
c) (4x - 8)(x2 + 6) - (4x - 8)(x + 7) + 9(8 - 4x)
= (4x - 8)(x2 + 6) - (4x - 8)(x + 7) - 9(4x - 8)
= (4x - 8)(x2 + 6 - x - 7 - 9)
= 2(x - 4)(x2 - x - 10)
a) x6 – x4 + 2x3 + 2x2
= x2(x4 – x2 + 2x + 2)
= x2[x2(x2 – 1) + 2(x + 1)]
= x2. [x2.(x -1).(x + 1) + 2(x+ 1)]
= x2 (x+ 1).[x2(x- 1)+ 2]
= x2(x + 1)(x3 – x2 + 2)
= x2(x + 1)[(x3 + 1) – (x2 – 1)]
= x2(x + 1).[(x + 1).(x2 – x + 1) - (x - 1).(x + 1)]
= x2(x + 1)(x + 1)( x2 – x + 1 – x + 1)
= x2(x + 1)2(x2 – 2x + 2).
b) 4x4 + y4 = 4x4 + 4x2y2 + y4 - 4x2y2
= (2x2 + y2)2 - (2xy)2
= (2x2 + y2 + 2xy)(2x2 + y2 - 2xy)