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a) 16x2 - ( x2 + 4 )2
= ( 4x )2 - ( x2 + 4 )2
= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]
= ( -x2 + 4x - 4 )( x2 + 4x + 4 )
= [ -( x2 - 4x + 4 ) ]( x + 2 )2
= [ -( x - 2 )2 ]( x + 2 )2
b) ( x + y )3 + ( x - y )3
= [ ( x + y ) + ( x - y ) ][ ( x + y )2 - ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]
= 2x( 2x2 + 2y2 - x2 + y2
= 2x( x2 + 3y2 )
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
a/\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(a^3-3a^2b+3ab^2-b^3\right)\)\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^2\)
\(=6ab^2+2b^3\)(rút gọn hết)
b/\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-2xz+2xz+2xy-3xz-3yz-3xy\right).\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Hok tốt
a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36y^2x^2-8y^3\)
c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)
a) \(\left(x^2+2xy\right)^3\)
\(=\left(x^2\right)^3+3\left(x^2\right)^22xy+3x^2\left(2xy\right)^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b) \(\left(3x^2-2y\right)^3\)
\(=\left(3x^2\right)^3-3\left(3x^2\right)^22y+3.3x^2\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
c) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\left(2x^3\right)^2y^2+3.2x^3\left(y^2\right)^2-\left(y^2\right)^3\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6.\)
Giải:
a) \(\left(2x+y+3\right)^2\)
\(=\left(2x+y\right)^2+2.3\left(2x+y\right)+3^2\)
\(=\left(2x\right)^2+2.2x.y+y^2+2.3\left(2x+y\right)+3^2\)
\(=4x^2+4xy+y^2+12x+6y+9\)
Vậy ...
b) \(\left(x-2y+1\right)^2\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1^2\)
\(=x^2-2.x.2y+\left(2y\right)^2+2x-4y+1^2\)
\(=x^2-4xy+4y^2+2x-4y+1\)
Vậy ...
c) \(\left(x^2-2xy^2-3\right)^2\)
\(=\left(x^2-2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=\left(x^2\right)^2-2.x^2.2xy^2+\left(2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=x^4-4x^3y^2+4x^2y^4+6x^2-12xy^2-9\)
Vậy ...
Trả lời:
a, \(\left(3\sqrt{x}-y\right)\left(3\sqrt{x}+y\right)=\left(3\sqrt{x}\right)^2-y^2=9x-y^2\)
b, \(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{y}+\sqrt{x}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+2\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(2\sqrt{y}\right)^2\)
\(=x-4y\)
\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)
Y = a^3 - b^3
Y = ( a - b ) ( a^2 + a . b + b^2 )