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14 tháng 9 2020

Bài 1.

a) ( 3x + 4y )2 = ( 3x )2 + 2.3x.4y + ( 4y )2 = 9x2 + 24xy + 16y2

b) ( x2 + 1 )2 = ( x2 )2 + 2.x2.1 + 12 = x4 + 2x2 + 1

c) ( 3 - 2y )2 = 32 - 2.3.2y + ( 2y )2 = 9 - 12y + 4y2

d) ( xy2 - 2 )2 = ( xy2 )2 - 2.xy2.2 + 22 = x2y4 - 4xy2 + 4

Bài 2.

a) x2 - 9 = x2 - 32 = ( x - 3 )( x + 3 )

b) 25 - 4y2 = 52 - ( 2y )2 = ( 5 - 2y )( 5 + 2y )

c) 9x4 - 4y2 = ( 3x2 )2 - ( 2y )2 = ( 3x2 - 2y )( 3x2 + 2y )

d) ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )

14 tháng 9 2020

B1:

a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)

b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)

c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)

d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)

B2:

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

b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)

c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)

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

19 tháng 12 2023

a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
 => (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)

k) (x + 5)3

27 tháng 8 2023

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

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

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

\(=\left(3x-2y\right)\left[3x+2y-1\right]\)

d) \(4x^2-4xy+2x-y+y^2\)

\(=4x^2-4xy+y^2+2x-y\)

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

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

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

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

27 tháng 8 2023

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

10 tháng 6 2015

B1)9x4+16y6-24x2y3=(3x2-4y3)2

B2)a)81-x4=(9-x2)(9+x2)=(3-x)(3+x)(9+x2)

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

c)(+y+z)2-(x-y-z)2=(x+y+z-x+y+z)(x+y+z+x-y-z)=(2y+2z)2x=4x(y+z)

B3)

(123+1)(123-1)-36.46

=126-1-(3.4)6

=126-1-126=-1

16 tháng 9 2021

\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)

23 tháng 7 2016

Toán lớp 8

23 tháng 7 2016

1. \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)  

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

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

4.  \(8+x^3=2^3+x^3=2^3+3.2^2.x+3.2.x^2+x^3\) 

                                   \(=8+12x+6x^2+x^3\)  

5. \(8x^3+27=\left(2x\right)^3+3^3\)  

                    \(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)  

                    \(=8x^3+36x^2+54x+27\)  

6. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\) 

7. \(\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\) 

8. \(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc\) 

AH
Akai Haruma
Giáo viên
13 tháng 11 2023

Yêu cầu đề là gì vậy bạn?

31 tháng 10 2021

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

1 tháng 11 2021

Bn cs lm đc câu c, d lun k

a: \(50x^5-8x^3\)

\(=2x^3\left(25x^2-4\right)\)

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

b: \(x^4-5x^2-4y^2+10y\)

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

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

c: \(36a^2+12a+1-b^2\)

\(=\left(6a+1\right)^2-b^2\)

\(=\left(6a+1-b\right)\left(6a+1+b\right)\)

d: \(x^3+y^3-xy^2-x^2y\)

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

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

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

e: Ta có: \(4x^2+4x-3\)

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

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

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

f: Ta có: \(9x^4+16x^2-4\)

\(=9x^4+18x^2-2x^2-4\)

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

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

g: Ta có: \(-6x^2+5xy+4y^2\)

\(=-6x^2+8xy-3xy+4y^2\)

\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)

\(=\left(3x-4y\right)\left(-2x-y\right)\)

h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)

\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)

\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)

a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)