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\(a,\frac{15x^2y^4}{5x^3z}=\frac{3y^4}{x}\)
\(b,\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
\(c,\frac{5x^2+10xy+5y^2}{15x+15y}=\frac{5\left(x^2+2xy+y^2\right)}{15\left(x+y\right)}=\frac{5\left(x+y\right)^2}{15\left(x+y\right)}=\frac{x+y}{3}\)
\(d,\frac{2x^3-2}{11x^2-22x+11}=\frac{2\left(x^3-1\right)}{11\left(x^2-2x+1\right)}=\frac{2\left(x-1\right)\left(x^2+x+1\right)}{11\left(x-1\right)^2}=\frac{2\left(x^2+x+1\right)}{11\left(x-1\right)}\)
= 5. (x^2 -2xy +y^2 - 4z^2)
= 5. (x^2 - 2xy +y^2) - 4z^2
= 5. (x - y)^2 - (2z)^2
= 5. [(x - y) - 2z].[(x - y) + 2z]
=5. (x - y - 2z). (x - y + 2z)
A = ( x + y )2 = 5 ( x - y ) + 1
A = 9 = 5( x - y ) + 1
A = 8 = 5 ( x - y )
A = 1,6 = x + y
=> A = 1,6
chắc sai
cứ tham khảo
5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x2 - 2xy + y2) - 4z2]
= 5 [(x - y)2 - (2z)2]
= 5 (x - y - 2z) (x - y +2z)
a )
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
b )
\(-5x^2-16x-3\)
\(=-5x^2-15x-x-3\)
\(=-5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(-5x-1\right)\left(x+3\right)\)
c )
\(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)-5\right]\)
d )
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
P/s : Mình bổ sung :
a )
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d )
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5x2 - 10xy + 5y2 - 20z = (5x2 + 5y2) - (10xy + 20z)
= 5 (x2 + y2) - 10 (xy + 2z)
= 5 [(x2 + y2) - 2 (xy + 2z)]
= 5 [(x2 + y2) - 2xy - 4z]
= -5 (4z - y2 + 2xy - x2)
Ta có: 5( x^2 +2xy +y^2 ) = 5(x+y)^2