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27 tháng 7 2016

X2+4xy-21y2=(x2+4xy+4y2)-25y2=(x+2)2-(5y)2=(x+2-5y)(x+2+5y)

5x2+6xy+y2=9x2+6xy+y2-4x2=(3x+y)2-4x2=(3x+y+2x)(3x+y-2x)

(x-y)2+4(x-y)-12=(x-y+2)2-16=(x-y+2+4)(x-y+2-4)

x2-7xy+10y2=x2-7xy+\(\frac{49y^2}{4}-\frac{9y^2}{4}\)\(\left(x-\frac{7}{2}\right)^2-\left(\frac{3y}{2}\right)^2\)=\(\left(x-\frac{7}{2}-\frac{3y}{2}\right)\left(x-\frac{7}{2}+\frac{3y}{2}\right)\)

x2+2xy-15y2=(x+y)2-16y2=(x+y-4y)(x+y+4y

22 tháng 10 2023

a: x^2+4xy-21y^2

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

=5x(x+y)+y(x+y)

=(x+y)(5x+y)

c: \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

=x(x+5y)-3y(x+5y)

=(x+5y)(x-3y)

d: \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

=x(x-2y)-5y(x-2y)

=(x-2y)(x-5y)

22 tháng 10 2023

a) \(x^2+4xy-21y^2\)

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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

c) \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

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

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

d) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-5y\right)\left(x-2y\right)\)

8 tháng 9 2023

a, \(x^2\) + 4\(x\) - y2 + 4

= (\(x^2\) + 4\(x\) + 4) - y2

= (\(x\) + 2)2 - y2

= (\(x\) + 2 - y)(\(x\) + 2 + y)

8 tháng 9 2023

b, 2\(x^2\) - 18

= 2.(\(x^2\) -9)

= 2.(\(x\) -3).(\(x\) + 3)

 

27 tháng 8 2021

c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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

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

c: \(5x^2+15x+3y+xy\)

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

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

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

\(=\left(x-y\right)^2-9\)

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

10 tháng 10 2021

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

10 tháng 10 2021

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

AH
Akai Haruma
Giáo viên
29 tháng 12 2023

Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.

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

b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$

c. $5x^2-10xy=5x(x-2y)$

d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$

$=(x-y)(x+y+1)$

21 tháng 12 2022

bạn tách ra từng ít câu 1 thôi ạ

22 tháng 12 2022

a: \(=5x\left(xy^2+3x+6y^2\right)\)

b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)

c: \(=\left(x-3\right)\left(x-4\right)\)

d: \(=x\left(x^2-2xy+y^2-9\right)\)

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

e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

f: \(=\left(x-4\right)\left(x+3\right)\)

 

a: Ta có: \(x^2-4y^2-2x-4y\)

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

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

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

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

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

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

e: Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

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

f: Ta có: \(x^2-y^2-2x-2y\)

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

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

20 tháng 10 2017

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

11 tháng 12 2021

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)