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

học căn chx

a: \(25x^2-\dfrac{10}{3}xy+\dfrac{1}{9}y^2=\left(5x-\dfrac{1}{3}y\right)^2\)

b: \(25x^2-15x+\dfrac{9}{4}=\left(5x-\dfrac{3}{2}\right)^2\)

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

d: \(\left(x^2-\dfrac{2}{3}\right)\left(x^4+\dfrac{2}{3}x^2+\dfrac{4}{9}\right)=x^6-\dfrac{8}{27}\)

B1:

\(=x^2+2x-5x-10+3\left(x^2-2^2\right)-\left(9x^2-2.3x.\frac{1}{2}+\frac{1}{4}\right)+5x^2\)

\(=-10-12-\frac{1}{4}=-22\frac{1}{4}\)

Bài 1.

( x - 5 )( x + 2 ) + 3( x - 2 )( x + 2 ) - ( 3x - 1/2 )² + 5x²

= x² - 3x - 10 + 3( x² - 4 ) - ( 9x² - 3x + 1/4 ) + 5x²

= 6x^2 _-- 3x - 10 + 3x² - 12 - 9x² + 3x - 1/4

= -89/4 không phụ thuộc vào biến

=> đpcm

Bài 2 < mình viết luôn nhé >

a) ( x + 2y² )² = x² + 4xy² + 4y⁴

b) ( a - 5/2b )² = a² - 5ab + 25/4b²

c) ( m + 1/2 )² = m² + m + 1/4

d) x² - 16y⁴ = ( x + 4y² )( x - 4y² )

e) 25a² - 1/4b² = ( 5a + 1/2b )( 5a - 1/2b )

a,2²;x;2

b,12x;3x;2

c,(1/2)²;x;1/2

d,?

a) \(x^2+4x+4=\left(x+2\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

c) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

Ta có \(a>b\)\(=>a+4>b+4\)

Nên bất đẳng thức b, là đúng

a) (x+2y)²=x²+4xy+4y²

^b) (a-3b)²=a²- 6ab+9b²

bài 1 : điền vào chỗ chấm để đk khẳng định đúng :

a) (.x..+2y...)²=x²+..4y.+4y²

b) (.a..-.3b..)²=a²-6ab+.9b²..

c) (.m..+.\(\frac{1}{2}\)..)²=.m²..+m+1/4

d) 25a²-..\(\frac{1}{4}b\).=(.5a..+1/2b)(..5a..-1/2b)

e)(.2x...+.1..)^2 = 4x^2 +.4x..+1

g)(2-x)(.4..+.2x..+.x²..)=8-x^3

h) 16a^2 - ..9. = (..4a.+3)(..4a.-3)

f)25 - ..30y.+9y^2=(..5.+...3y.)^2

bài 4 : tính :

a) (x+1)(x^2-x+1)= x³+ 1

b) (x-2)(x^2+2x+4)= x³- 2

c)(1-x)(1+x+x^2)= 1 - x³

d) a^3+3a^2+3a+1 = (a + 1 )³

e) b^3-6b^2+12b-8 = (b - 2 )³

A=(a+1)(a+2)(a^2+4)(a-1)(a^2+1)(a-2)

A =(a+1)(a-1)(a+2)(a-2)(a^2+4)(a^2+1)

A =(a^2-1)(a^2+1)(a^2-4)(a^2+4)

A =(a^4-1)(a^4-16)

A =\(a^{16}-16\cdot a^4-a^4+16\)

A =\(a^{16}-17\cdot a^4+16\)

B=(a+2b-3c-d)(a+2b+3c+d)

B=[(a+2b)^2 - (3c +d)^2]

B=[a^2+4ab+4b^2-(9c^2+6cd+d^2)]

B=a^3+4ab+4b^2 - 9c^2 - 6cd - d^2

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

C=[(1-x)^2-(2x^3-3x^2)^2]

C=[(1-2x+x^2) - (4x^6-12x^5+9x^4)]

C=[1-2x-x^2-4x^6+12x^5-9x^4]

C=-4x^6+12x^5-9x^4-x^2-2x+1

D=(a^6-3a^3+9)(a^3+3)

D=a^9+27

a, bằng cách tìm nhân tử chung

1,\(x^2-3x\)

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

2,\(15x^2-6x\)

=3x.(5x-2)

3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)

=(x-y).(4x+2y)

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

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

b, dùng hằng đẳng thức

1,\(64x^2-25y^2\)

=\(\left(8x\right)^2-\left(5y\right)^2\)

=(8x-5y)(8x+5y)

2,\(9x^2-30x-25\)

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

3,

\(\dfrac{1}{4}x^2+2x+4\)

=\(\left(\dfrac{1}{2}x+2\right)^2\)

4,\(25a^2-2a+\dfrac{1}{25}\)

=(\(\left(5a-\dfrac{1}{5}\right)^2\)

Phân tích hằng đẳng thức

a) Bằng cách tìm nhân tử chung

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

2) \(15x^2-6x=3x\left(5x-2\right)\)

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

4) \(5\left(a-2b\right)-10x\left(b-2a\right)=5a-10b-10xb+20xa=5\left(a-2b-2xb+4xa\right)\)

b) Bằng cách dùng hằng đẳng thức

1) \(64x^2-25y^2=\left(8x-5y\right)\left(8x+5y\right)\)

2) \(9x^2-30x+25=\left(3x\right)^2-2.3x.5+5^2=\left(3x-5\right)^2\)

3) \(\dfrac{1}{4}x^2+x+4=\left(\dfrac{1}{2}x\right)^2+\dfrac{1}{2}.x.2+2^2=\left(\dfrac{1}{2}x+2\right)^2\)

4) \(25a^2-2a+\dfrac{1}{25}=\left(5a\right)^2-5.2.a+\left(\dfrac{1}{5}\right)^2=\left(5a-\dfrac{1}{5}\right)^2\)