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\(A=5x^3-125x=5x\left(x-5\right)\left(x+5\right)\)
\(B=x^3-8+\left(x-2\right)\left(5x+4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4+5x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x+4\right)\)
a, \(100a^2-\left(a^2+25\right)^2=\left(10a\right)^2-\left(a^2+25\right)^2\)
\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)=-\left(a-5\right)^2\left(a+5\right)^2\)
b,\(-5\left(xy\right)^3-5\left(xy\right)^3=-10\left(xy\right)^3\)
c,\(16+2\left(xy\right)^3=2\left(2+xy\right)\left(4-2xy+x^2y^2\right)\)
\(=x^2y\left(x-5\right)-2y\left(x-5\right)+0\)
\(=\left(x-5\right)\left(x^2y-2y\right)\)
xong phân tích nốt cái bậc 2 kia để được max điểm :)))
\(xy^3-5x^2\)
\(=x\cdot y^3-x\cdot5x\)
\(=x\left(y^3-5x\right)\)
\(y^2+3y\)
\(=y\cdot y+y\cdot3\)
\(=y\left(y+3\right)\)
\(27y^3-9y^2+y-\frac{1}{27}\)
\(=\left(3y\right)^3-3.\left(3y\right)^2.\frac{1}{3}+3.3y.\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^3\)
\(=\left(3y-\frac{1}{3}\right)^3\)
hk
tốt
a) xz-yz+5y-5x=\(z\left(x-y\right)+5\left(y-x\right)\)=\(z\left(x-y\right)-5\left(x-y\right)\)=\(\left(z-5\right)\left(x-y\right)\)
b) \(3x^2-6x+3-3y^2\)=\(3\left(x^2-2x+1-y^2\right)\)=\(3\left(\left(x-1\right)^2-y^2\right)\)=\(3\left(x-1-y\right)\left(x-1+y\right)\)
\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-27y^3\)
\(=\left(x+1-3y\right)\left(x^2+3xy+9y^2\right)\)
** Sửa: $125x^3-5x-3y+27y^3$
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Lời giải:
$125x^3-5x-3y+27y^3=(125x^3+27y^3)-(5x+3y)$
$=[(5x)^3+(3y)^3]-(5x+3y)$
$=(5x+3y)(25x^2-15xy+9y^2)-(5x+3y)$
$=(5x+3y)(25x^2-15xy+9y^2-1)$