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a: \(=\left(-a^2-2a+3\right)^2\)

b: \(=\left(a^2+3\right)^2-4a^2\)

c: \(=-\left(a^2-2a\right)\left(a^2+2a\right)=-\left(a^4-4a^2\right)\)

a) Áp dụng hằng đẳng thức : \(a^2-b^2+\left(a-b\right)\left(a+b\right)\)

Ta có ; \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left[\left(a^2+2a\right)+3\right]\left[\left(a^2+2a\right)-3\right]\)

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

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

1:

a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)

b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)

c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)

2 tháng 9 2021

Bài 2: tất cả đều ở dạng tích rồi mà

Bài 1:

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

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

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

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

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

g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)

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

Bài 2 :

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

b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)

c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)

e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)

29 tháng 8 2020

( a2 - 2a + 3 )( a2 + 2a - 3 )

= [ a2 - ( 2a - 3 ) ][ a2 + ( 2a - 3 ) ]

= ( a2 )2 - ( 2a - 3 )2

= a4 - ( 4a2 -  12a + 9 )

= a4 - 4a2 + 12a - 9 

29 tháng 8 2020

\(\left(a^2-2a+3\right)\left(a^2+2a-3\right)\)

\(=a^4+2a^3-3a^2-2a^3-4a^2+6a+3a^2+6a-9\)

\(=a^4-4a^2+12a-9\)

a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)

\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)

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

\(=a^4-\left(2a+3\right)^2\)

b: \(\left(-a^2-2a+3\right)^2\)

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

\(=a^4+4a^2+9+4a^3-18a-6a^2\)

\(=a^4+4a^3-2a^2-18a+9\)

c: \(\left(x-y-z\right)^2\)

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

\(=x^2-2xy-2xz+y^2+2yz+z^2\)

d: \(\left(x+y+z\right)\left(x-y-z\right)\)

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

\(=x^2-y^2-2yz-z^2\)

a: \(=a^2-b^4\)

b: \(=\left(a^2+2a\right)^2-9\)

c: \(=a^2-\left(2a+3\right)^2\)

d: \(=a^4-\left(2a-3\right)^2\)

e: \(=\left(-a^2-2a+3\right)^2\)

g: \(=4a^2-a^4\)

12 tháng 7 2018

\(a^22\) là a2 nhân với 2 đó hả?

\(a,\left(a^22+2a+3\right)\left(a^22+2a-3\right)\)

\(=\left[\left(a^22+2a\right)+3\right]\left[\left(a^22+2a\right)-3\right]\)

\(=\left(a^22+2a\right)^2-9\)

\(=4a^4+8a^3+4a^2-9\)

\(b,\left(a^22+2a+3\right)\left(a^2-2a-3\right)\)

\(=2a^4-4a^3-6a^2+2a^3-4a^2-6a+3a^2-6a-9\)

\(=2a^4-2a^3-7a^2-12a-9\)

\(c,\left(a^22-2a+3\right)\left(a^2+2a-3\right)\)

\(=2a^4+4a^3-6a^2-2a^3-4a^2+6a+3a^2+6a-9\)

\(=2a^4+2a^3-7a^2+12a-9\)