Ai trả lời mình đi cho đỡ quê 😞💔

a: \(=\left(x+2-y\right)\left(x+2+y\right)\)

c: \(=\left(x-y\right)^2\)

a) \(x^2+4x-y^2+4\)

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

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

c) \(x^2-2xy+y^2-z^2+2zt-t^2\)

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

\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

a) x² + 4x - y² + 4 = (x² + 4x + 4) - y² = (x + 2)² - y = (x + 2 - y)(x + 2 + y)

b) 3x² + 6xy + 3y² - 3z² = 3(x² + 2xy + y² - z²) = 3[(x² + 2xy + y²) - z²) = 3[(x + y)² - z) = 3(x + y + z)(x + y - z)²

a) x² + 4x – y² + 4;                    

=x²+4x+4-y²

=(x+2)²-y²

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

b) 3x² + 6xy + 3y² – 3z²;

=3.(x²+2xy+y²)-3z²

=3.(x+y)²-3z²

=3.[(x+y)²-z²]

=3.(x+y-x)(x+y+z)

c) x² – 2xy + y² – z² + 2zt – t².

=(x-y)²-(z²-2zt+t²)

=(x-y)²-(z-t)²

=[(x-y)-(z-t)][(x-y)+(z-t)]

=(x-y-z+t)(x-y+z-t)

a) Ta có : x² + 4x – y² + 4

= x² + 4x + 4 - y² 

= (x + 2)² - y²

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

b) 3x² + 6xy + 3y² - 3z² 

= 3(x² + 2xy + y²) - 3z²

= 3(x + y)² - 3z²

= 3[(x + y)² - z²]

= 3(x + y - z)(x + y + z)

\(x^2+4x-y^2+4\)

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

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

hk tốt

^^

a) x² + 4x – y² + 4;                    

=x²+4x+4-y²

=(x+2)²-y²

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

b) 3x² + 6xy + 3y² – 3z²;

=3.(x²+2xy+y²)-3z²

=3.(x+y)²-3z²

=3.[(x+y)²-z²]

=3.(x+y-x)(x+y+z)

c) x² – 2xy + y² – z² + 2zt – t².

=(x-y)²-(z²-2zt+t²)

=(x-y)²-(z-t)²

=[(x-y)-(z-t)][(x-y)+(z-t)]

=(x-y-z+t)(x-y+z-t)

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

b; \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)

c, \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t^2\right)=\left(x-y-z+t\right)\left(x+y+z-t\right)\)

quas ngu

a, 3( x² +2xy + y² - z²) = 3((x+y)² - z²) = 3( x+y-z)(x+y+z)

b, (x-y)² - (z-t)² = (x-y+z-t)(x-y-z+t)

a/3x^2+6xy+3y^2-3z^2

=3(x^2+2xy+y^2-z^2)

=3[(x+y)^2-x^2)]

=3(x+y-z)(x+y+z)

b/x^2-2xy+y^2-z^2+2zt-t^2

=(x-y)^2-(z-t)^2

=(x-y+z-t)(x-y-z+t)

\(a,x^2+6x+9\)

\(=x^2+3x+3x+9\)

\(=\left(x^2+3x\right)+\left(3x+9\right)\)

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

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

\(=\left(x+3\right)^2\)

\(b,10x-25-x^2\)

\(=-\left(x^2-2.5.x+5^2\right)\)

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

\(c,x^2+4x-y^2+4\)

\(=\left(x^2+2.2.x+2^2\right)-y^2\)

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

\(=\left(x+2-y\right).\left(x+2+y\right)\)

\(d,3x^2+6xy+3y^2-3z^2\)

\(=3.[\left(x^2+2xy+y^2\right)-z^2]\)

\(=3.[\left(x+y\right)^2-z^2]\)

\(=3.\left(x+y-z\right)\left(x+y+z\right)\)

\(e,x^2-2xy+y^2-z^2+2zt-t^2\)

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

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

\(=[\left(x-y\right)-\left(z-t\right)].[\left(x-y\right)+\left(z-t\right)]\)

\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)

bai tim x bai 5 co 

\(x^6+6x+9\)

\(=x^2+2.x.3+3^2\)

\(=\left(x+3\right)^2\)

hk tốt

^^

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

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

b) \(3x^2+6xy+3y^2-3z^2\Leftrightarrow\left(\sqrt{3}x+\sqrt{3}y\right)^2-\left(\sqrt{3}z\right)^2\)

\(\Leftrightarrow\left(\sqrt{3}x+\sqrt{3}y-\sqrt{3}z\right).\left(\sqrt{3}x+\sqrt{3}y+\sqrt{3}z\right)\)

c) \(x^2-2xy+y^2-z^2+2zt-t^2\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

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