1/ tính nhẩm theo các hằng đẳng thức các số sau:
a) 19^2;28^2;81^2;91^2
b)19.21;29.31;39.41
c)29^2-8^2;56^2-46^2;67^2-56^2
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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 !
\(19^2=\left(20-1\right)^2=20^2-2.20.1+1^2=400-40+1=361\)
\(28^2=\left(30-2\right)^2=30^2-2.30.2+2^2=900-120+4=784\)
\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)
\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)
\(19.21=\left(20-1\right).\left(20+1\right)=20^2-1^2=400-1=399\)
\(29.31=\left(30-1\right).\left(30+1\right)=30^2-1^2=900-1=899\)
\(39.41=\left(40-1\right).\left(40+1\right)=40^2-1^1=1600-1=1599\)
\(29^2-8^2=\left(29-8\right).\left(29+8\right)=777\)
- 2 phần còn lại bạn cứ làm tương tự :) Vì mk bận nên chỉ giúp đc đến đây thoy <: Chúc bạn học tốt =)
\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)
\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)
a)
\(19^2=\left(20-1\right)^2=20^2-2.20.1+1^2=400-40+1=361\)
\(28^2=\left(30-2\right)^2=30^2-2.30.2+2^2=900-120+4=784\)
\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)
\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)
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) \(=4x^2-12x+9\)
b) \(=4x^2+2x+\dfrac{1}{4}\)
c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)
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)