cho a 1

L.I.K.E

để a 

làm hộ bn này bài này nào 

\(19^2=\left(20-1\right)^2=20^2-2.20.1+1^2=400-40+1=361\)

\(28^2=\left(30-2\right)^2=30^2-2.30.2+2^2=900-120+4=784\)

\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)

\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)

\(19.21=\left(20-1\right).\left(20+1\right)=20^2-1^2=400-1=399\)

\(29.31=\left(30-1\right).\left(30+1\right)=30^2-1^2=900-1=899\)

\(39.41=\left(40-1\right).\left(40+1\right)=40^2-1^1=1600-1=1599\)

\(29^2-8^2=\left(29-8\right).\left(29+8\right)=777\)

- 2 phần còn lại bạn cứ làm tương tự :) Vì mk bận nên chỉ giúp đc đến đây thoy <: Chúc bạn học tốt =)

C

a) `(3/5 x^2 -1/2 y)^2 = (3/5 x^2)^2 - 2. 3/5 x^2 .1/2 y + (1/2 y)^2`

`= 9/25 x^4 - 3/5 x^2y + 1/4 y^2`

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

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

5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

4:

a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2

b: 8(7^2+1)(7^4+1)(7^8+1)

=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^16-1)<7^16-1

5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

mik chỉ biết bài 5 thôi !

a)

\(19^2=\left(20-1\right)^2=20^2-2.20.1+1^2=400-40+1=361\)

\(28^2=\left(30-2\right)^2=30^2-2.30.2+2^2=900-120+4=784\)

\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)

\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)

b)

\(19.21=\left(20-1\right)\left(20+1\right)=20^2-1^2=400-1=399\)

\(29.31=\left(30-1\right)\left(30+1\right)=30^2-1^2=900-1=899\)

\(39.41=\left(40-1\right)\left(40+1\right)=40^2-1^2=1600-1=1599\)

P/s: Lần sau cậu nên chia nhỏ ra đăng nhé!

\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)

\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)

a) (2xy)²-2(2xy-3)+3²

a)\(\left(2xy-3\right)^2=\left(2xy\right)^2-2\cdot2xy\cdot3+3^2=4x^2y^2-12xy+9\)

b)\(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot\dfrac{1}{3}y+\left(\dfrac{1}{3}y\right)^2\)

                          \(=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2\cdot a\cdot b+b^2\)

\(=a^2-2ab+b^2\)

\(=a^2-4ab+2ab+b^2\)

\(=\left(a^2+2ab+b^2\right)-4ab\)

\(=\left(a+b\right)^2-4ab=VP\)

⇒ Đpcm

b) Ta có:

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

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

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

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

\(=2x^2+0+2y^2\)

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

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

⇒ Đpcm

a: (a-b)^2

=a^2-2ab+b^2

=a^2+2ab+b^2-4ab

=(a+b)^2-4ab

b: (x+y)^2+(x-y)^2

=x^2+2xy+y^2+x^2-2xy+y^2

=2x^2+2y^2

=2(x^2+y^2)

a) \(=4x^2-12x+9\)

b) \(=4x^2+2x+\dfrac{1}{4}\)

c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

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

e) \(=\left(3-\dfrac{x}{2}\right)\left(9+\dfrac{3x}{2}+\dfrac{x^2}{4}\right)\)

f) \(=\left(125-4x\right)\left(125^2+500x+16x^2\right)\)