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\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
Đề bài không chính xác, biểu thức này không viết được dưới dạnh tích
\(27-x^3=3^3-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)
`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`
`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`
`=-(x/2 - 1)^3`
`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`
`=(x^3 - 1/(2y))^{3}`
\(\frac{1}{27}+x^3\)
\(=\left(\frac{1}{3}\right)^3+x^3\)
\(=\left(\frac{1}{3}+x\right)\left(\frac{1}{9}-\frac{1}{3}x+x^2\right)\)
\(\frac{1}{27}+x^3=\left(\frac{1}{3}\right)^3+x^3\)\(\left(\frac{1}{3}+x\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}x+x^2\right]\)
\(x^3-\dfrac{1}{8}\\ =x^3-\left(\dfrac{1}{2}\right)^3\\ =\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)