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26 tháng 11 2016

\(\left(x+1\right)^2-25=\left(x+26\right)\left(x-24\right)\)

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

26 tháng 11 2016

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

\(=x^2+2x+1-25\)

\(=x^2+2x-24\)

\(=x^2+6x-4x-24\)

\(=x\left(x+6\right)-4\left(x+6\right)\)

\(=\left(x-4\right)\left(x+6\right)\)

\(1-2y+y^2\)

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

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

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

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

C2

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

11 tháng 11 2021

từng câu 1 thôi:v

 

11 tháng 11 2021

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

26 tháng 11 2021

\(a,=5\left(a-4b\right)\\ b,=\left(y+1\right)^2-x^2=\left(y+1-x\right)\left(x+y+1\right)\)

26 tháng 11 2021

a) 5a - 20b

= 5 ( a - 4b )

b) y^2 + 2y - x^2 + 1

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

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

= ( y + 1 + x ) ( y + 1 - x )

e) Ta có: \(x^4-2x^3+2x-1\)

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

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

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

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

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

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

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

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

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

a) Ta có: \(x^2-y^2-2x-2y\)

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

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

b) Ta có: \(x^2\left(x+2y\right)-x-2y\)

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

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

Câu 1: A

Câu 21: A

 

1 tháng 11 2021

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

27 tháng 9 2021

a. (2x - 5)2

b. x(x2 - 1) = x(x - 1)(x + 1)

c. (x - 3)(x2 + 3x + 9)

d. 5x(x - y) + (x - y) = (x - y)(5x + 1)

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

a. 

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

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

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

d.

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

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

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

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

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$

$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$

$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

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

30 tháng 7 2023

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

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

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

\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)

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

c: \(=x^2-8x+x-8\)

\(=x\left(x-8\right)+\left(x-8\right)\)

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

9 tháng 10 2023

\(a,xy+y^2-x-y\)

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

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

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

\(---\)

\(b,\left(x^2y^2-8\right)^2-1\)

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

\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)

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

\(---\)

\(c,x^2-7x-8\)

\(=x^2+x-8x-8\)

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

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

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

\(Toru\)

NV
10 tháng 1 2022

Kiểm tra lại \(x^2-2y+y^2...\) hay \(x^2-2xy+y^2...\)

a: 3x^2-9

=3*x^2-3*3

=3(x^2-3)

b: 1/2x^2-2y^2

=1/2(x^2-4y^2)

=1/2(x-2y)(x+2y)

c: 3x^2-12y^2

=3(x^2-4y^2)

=3(x-2y)(x+2y)

d: 1/3x^2y^2-3x^2

=1/3x^2(y^2-9)

=1/3x^2(y-3)(y+3)