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\(a,4x^4-21x^2y^2+y^4=\left(2x^2\right)^2+4x^2y^2+y^4-4x^2y^2-21x^2y^2\)

\(=\left(2x^2+y^2\right)^2-25x^2y^2\)

\(=\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)

\(b,x^5-5x^3+4x=x\left(x^4-5x^2+4\right)\)

\(=x\left(x^4-4x^2-x^2+4\right)\)

\(=x\left[x^2\left(x^2-4\right)-\left(x^2-4\right)\right]\)

\(=x\left(x^2-4\right)\left(x^2-1\right)\)

\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

\(c,x^3+5x^2+3x-9=x^3-x^2+6x^2-6x+9x-9\)

\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+6x+9\right)\)

\(=\left(x-1\right)\left(x^2+3x+3x+9\right)\)

\(=\left(x-1\right)\left[x\left(x+3\right)+3\left(x+3\right)\right]\)

\(=\left(x-1\right)\left(x+3\right)\left(x+3\right)\)

\(=\left(x-1\right)\left(x+3\right)^2\)

\(d,x^{16}+x^8-2=x^{16}+2x^8-x^8-2\)

\(=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)

\(=\left(x^8-1\right)\left(x^8+2\right)\)

AH
Akai Haruma
Giáo viên
7 tháng 9 2021

Lời giải:

a.

$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$

b.

$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$

$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$

$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$

$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$

c.

$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$

$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$

$=(x-1)(x^2+4x+7)$

7 tháng 9 2021

a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)

\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)

b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)

\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)

\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)

c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)

\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)

 

27 tháng 9 2023

a) \(x^4-y^4\)

\(=\left(x^2\right)^2-\left(y^2\right)^2\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

\(=\left(3x-2y+2x-3y\right)\left(3x-2y-3x+2y\right)\)

\(=0\cdot0\)

\(=0\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2\)

\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)

\(=\left(x-5y\right)\left(5x-y\right)\)

e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)

\(=\left(2x-1\right)^2-\left(x+1\right)^2\)

\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)

\(=3x\left(x-2\right)\)

f) \(x^3+27\)

\(=x^3+3^3\)

\(=\left(x+3\right)\left(x^2-3x+9\right)\)

g) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(0,1\right)^3\)

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

27 tháng 9 2023

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)

\(=\left(5x-5y\right)\left(x+y\right)\)

\(=5\left(x+y\right)\left(x-y\right)\)

1A. Phân tích các đa thức sau thành nhân tử:a) x3+2x;                                           b) 3x - 6y;c) 5(x + 3y)- 15x(x + 3y);               d) 3(x-y)- 5x(y-x).1B. Phân tích các đa thức sau thành nhân tử:a) 4x2 - 6x;                                         b) x3y - 2x2y2 + 5xy; c) 2x2(x +1) + 4x(x +1);               d) 2 x(y - 1) - 2 y(1 - y). 5                  52A. Phân tích các đa thức sau thành nhân tử: a) 2(x -1)3 - 5(x -1)2 - (x - 1);b) x(y - x)3 - y(x - y)2 + xy(x -...
Đọc tiếp

1A. Phân tích các đa thức sau thành nhân tử:

a) x3+2x;                                           b) 3x - 6y;

c) 5(x + 3y)- 15x(x + 3y);               d) 3(x-y)- 5x(y-x).

1B. Phân tích các đa thức sau thành nhân tử:

a) 4x2 - 6x;                                         b) x3y - 2x2y2 + 5xy;

 

c) 2x2(x +1) + 4x(x +1);               d) 2 x(y - 1) - 2

 

y(1 - y).

 

5                  5

2A. Phân tích các đa thức sau thành nhân tử: a) 2(x -1)3 - 5(x -1)2 - (x - 1);

b) x(y - x)3 - y(x - y)2 + xy(x - y);

c) xy(x + y)- 2x - 2y;

d) x(x + y)2 - y(x + y)2 + y2 (x - y).

2B. Phân tích đa thức thành nhân tử: a) 4(2-x)2 + xy - 2y;

b) x(x- y)3 - y(y - x)2 - y2(x - y);

c) x2y-xy2 - 3x + 3y;

d) x(x + y)2 - y(x + y) 2 + xy - x 2 .

3

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: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

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)

 

24 tháng 9 2021

\(a,=\left(3x+1\right)^2-y^2=\left(3x-y+1\right)\left(3x+y+1\right)\\ b,=x\left(x^2-5x+6\right)=x\left(x^2-2x-3x+6\right)=x\left(x-2\right)\left(x-3\right)\)

a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

b: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c: \(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)

\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)

 

26 tháng 8 2021

a)\(x^4+4\\ =\left(x^2\right)^2+4x^2+4-4x^2\\ =\left[\left(x^2\right)^2+4x^2+4\right]-\left(2x\right)^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ =\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)

 

a) Ta có: \(x^2-3x+xy-3y\)

\(=x\left(x-3\right)+y\left(x-3\right)\)

\(=\left(x-3\right)\left(x+y\right)\)

b) Ta có: \(x^3+10x^2+25x-xy^2\)

\(=x\left(x^2+10x+25-y^2\right)\)

\(=x\left(x+5-y\right)\left(x+5+y\right)\)

c) Ta có: \(x^3+2+3\left(x^3-2\right)\)

\(=4x^3-4\)

\(=4\left(x-1\right)\left(x^2+x+1\right)\)

NV
21 tháng 8 2021

\(9x^2-12xy-20y-25=9x^2-25-4y\left(3x+5\right)\)

\(=\left(3x+5\right)\left(3x-5\right)-4y\left(3x+5\right)=\left(3x+5\right)\left(3x-4y-5\right)\)

\(xy^2-49x^3-28x^2-4x=x\left[y^2-\left(49x^2+28x+4\right)\right]\)

\(=x\left[y^2-\left(7x+2\right)^2\right]=x\left(y+7x+2\right)\left(y-7x-2\right)\)

\(x^2-3x-2019.2022=x^2-3x-2019\left(2019+3\right)\)

\(=x^2-3x-2019^2-3.2019=\left(x-2019\right)\left(x+2019\right)-3\left(x+2019\right)\)

\(=\left(x+2019\right)\left(x-2022\right)\)

a: \(9x^2-12xy-20y-25\)

\(=\left(3x-5\right)\left(3x+5\right)-4y\left(3x+5\right)\)

\(=\left(3x+5\right)\left(3x-5-4y\right)\)

 

 

19 tháng 12 2021

\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)

\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)

31 tháng 12 2021

c: =(x-2)(x-4)

b: \(=x\left(x^2+2xy+y^2-4\right)\)

=x(x+y-2)(x+y+2)

31 tháng 7 2021

a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)

b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)

c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

\(=x^3-x+5x-5\)

\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+5\right)\)

b) Ta có: \(x^3-3x^2+4\)

\(=x^3+x^2-4x^2+4\)

\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-4x+4\right)\)

\(=\left(x+1\right)\cdot\left(x-2\right)^2\)

c) Ta có: \(x^3+2x^2+3x+2\)

\(=x^3+x^2+x^2+x+2x+2\)

\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+x+2\right)\)

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

\(=\left(x+y\right)^2+2\left(x+y\right)-3\)

\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)

\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)

\(=\left(x+y+3\right)\left(x+y-1\right)\)