15x^2 + 10xy
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\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)
a) 15x2y : ( - 5xy)
= -5xy( -3x) : ( - 5xy)
= -3x
b) ( 15x2y - 10xy2 + 5xy) : 5xy
= 5xy( 3x - 2y + 1) : 5xy
= 3x - 2y + 1
c) ( x +y)3 : ( x+y)
= (x + y)( x +y)2 : ( x+ y)
= ( x +y)2
1)\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\frac{2y}{5\left(x+y\right)^2}\)
2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)
3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)
4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)
\(a,\frac{15x^2y^4}{5x^3z}=\frac{3y^4}{x}\)
\(b,\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
\(c,\frac{5x^2+10xy+5y^2}{15x+15y}=\frac{5\left(x^2+2xy+y^2\right)}{15\left(x+y\right)}=\frac{5\left(x+y\right)^2}{15\left(x+y\right)}=\frac{x+y}{3}\)
\(d,\frac{2x^3-2}{11x^2-22x+11}=\frac{2\left(x^3-1\right)}{11\left(x^2-2x+1\right)}=\frac{2\left(x-1\right)\left(x^2+x+1\right)}{11\left(x-1\right)^2}=\frac{2\left(x^2+x+1\right)}{11\left(x-1\right)}\)
\(\frac{5x^2+10xy+5y^2}{3x^3+3y^3}=\frac{5\left(x^2+2xy+y^2\right)}{3\left(x^3+y^3\right)}=\frac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)
\(\frac{-15x\left(x-y\right)}{3\left(y-x\right)}=\frac{15x\left(y-x\right)}{3\left(y-x\right)}=\frac{15x}{3}\)
a,\(15x^3y^4-20x^4y^3+30x^3y^3\)
=\(5x^3y^3\left(3y-4x+6\right)\)
b,\(x^2+10xy+25y^2\)
=\(x^2+2.x.5.y+\left(5y\right)^2\)
=\(\left(x+5y\right)^2\)
c,\(x^2-2xy+y^2-9z^2\)
=\(\left(x^2-2xy+y^2\right)-\left(3z\right)^2\)
=\(\left(x-y\right)^2-\left(3z\right)^2\)
=\(\left(x-y+3z\right)\left(x-y-3z\right)\)
chúc bn hok tốt
a, x^3+2x^2=x^2(x+2)
b,5x^2y-10xy+20x^2yz=5xy(x-2+4xz)
c, 3x^3-12x^2+21x^4=3x^2(x-4+7x^2)
d,3*(x+5y)-15x*(x+5y)=3(x+5y)(1-5x)
e,2x*(x-y)-4y^2+4x^2=(x-y)[5(x+y)+x-y}=(x-y)(6x-4y)=2(x-y)(3x-2y)
a) Ta có: \(5y^3-10xy^2+5yx^2-20y\)
\(=5y\left(y^2-2xy+x^2-4y\right)\)
b) Ta có: \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\cdot\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
c) Ta có: \(9x^2+y^2+6xy\)
\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)
\(=\left(3x+y\right)^2\)
d) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
e) Ta có: \(125x^3-75x^2+15x-1\)
\(=\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot1+3\cdot5x\cdot1^2-1^3\)
\(=\left(5x-1\right)^3\)
mình ko biết đề là gì nhưng nếu phân tích đa thức thành nhân tử thì là
5x(3x + 2y)