phan tich da thuc thanh nhan tu
3x^3 +12x^2 +12x
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Lời giải:
$x^3-4x^2-12x+27$
$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$
$=x^2(x+3)-7x(x+3)+9(x+3)$
$=(x+3)(x^2-7x+9)$
\(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)
\(a,x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(b,10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(c,8x^3-\frac{1}{8}\)
\(=8x^3-\left(\frac{1}{2}\right)^3\)
\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)
\(d,8x^3+12x^2+6xy^2+y^3\)
\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)
hok tốt!
phan tich cac da thuc sau thanh nhan tu theo mau:
a)\(2x^3-x\)
\(=x\left(2x^2-1\right)\)
\(=x\left(\left(\sqrt{2}x\right)^2-1^2\right)\)\
\(=x\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\)
b)\(5x^2\left(x-1\right)-15x\left(x-1\right)\)
\(=\left(5x^2-15x\right)\left(x-1\right)\)
\(=5x\left(x-3\right)\left(x-1\right)\)
d)\(3x\left(x-2y\right)+6y\left(2y-x\right)\)
\(=3x\left(x-2y\right)-6y\left(x-2y\right)\)
\(=\left(3x-6y\right)\left(x-2y\right)\)
\(=3\left(x-2y\right)\left(x-2y\right)\)
\(=3\left(x-2y\right)^2\)
tu de bai suy ra: 3x(x^3+4x+4)=3x(x+2)^2
3x^3+12x^2+12x=3x(x^2+4x+4)=3x(x+4)^2